Find x in inverse trigonometry equation












2














Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$



How do I find $x$?



I tried solving by differentiating both sides, but I get $x=0$.



How do you solve it, purely using trigonometric techniques?






trigonometry inverse-function







share|cite|improve this question
























  • Possibly useful - math.stackexchange.com/questions/672575/…
    – Eevee Trainer
    1 hour ago










  • But I don't know x^2+y^2
    – user5722540
    1 hour ago










  • I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
    – D.B.
    12 mins ago












  • Online open ended answer. Deadline of 150 seconds.
    – user5722540
    9 mins ago
















2














Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$



How do I find $x$?



I tried solving by differentiating both sides, but I get $x=0$.



How do you solve it, purely using trigonometric techniques?






trigonometry inverse-function







share|cite|improve this question
























  • Possibly useful - math.stackexchange.com/questions/672575/…
    – Eevee Trainer
    1 hour ago










  • But I don't know x^2+y^2
    – user5722540
    1 hour ago










  • I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
    – D.B.
    12 mins ago












  • Online open ended answer. Deadline of 150 seconds.
    – user5722540
    9 mins ago














2












2








2







Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$



How do I find $x$?



I tried solving by differentiating both sides, but I get $x=0$.



How do you solve it, purely using trigonometric techniques?






trigonometry inverse-function







share|cite|improve this question















Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$



How do I find $x$?



I tried solving by differentiating both sides, but I get $x=0$.



How do you solve it, purely using trigonometric techniques?






trigonometry inverse-function




trigonometry inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Eevee Trainer

5,0271734




5,0271734










asked 1 hour ago









user5722540user5722540

1518




1518












  • Possibly useful - math.stackexchange.com/questions/672575/…
    – Eevee Trainer
    1 hour ago










  • But I don't know x^2+y^2
    – user5722540
    1 hour ago










  • I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
    – D.B.
    12 mins ago












  • Online open ended answer. Deadline of 150 seconds.
    – user5722540
    9 mins ago


















  • Possibly useful - math.stackexchange.com/questions/672575/…
    – Eevee Trainer
    1 hour ago










  • But I don't know x^2+y^2
    – user5722540
    1 hour ago










  • I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
    – D.B.
    12 mins ago












  • Online open ended answer. Deadline of 150 seconds.
    – user5722540
    9 mins ago
















Possibly useful - math.stackexchange.com/questions/672575/…
– Eevee Trainer
1 hour ago




Possibly useful - math.stackexchange.com/questions/672575/…
– Eevee Trainer
1 hour ago












But I don't know x^2+y^2
– user5722540
1 hour ago




But I don't know x^2+y^2
– user5722540
1 hour ago












I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
– D.B.
12 mins ago






I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
– D.B.
12 mins ago














Online open ended answer. Deadline of 150 seconds.
– user5722540
9 mins ago




Online open ended answer. Deadline of 150 seconds.
– user5722540
9 mins ago










5 Answers
5






active

oldest

votes


















2














There is a useful identity that we can use in this case:



$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$



From here we can substitute:



$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$



We are then left with:



$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$



From here, you can solve for $x$.






share|cite|improve this answer





















  • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
    – user5722540
    1 hour ago





















2














Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.






share|cite|improve this answer























  • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
    – user5722540
    59 mins ago



















1














I'm gonna derive the general function for $arcsin x$ then go from there.



Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found






share|cite|improve this answer





























    0














    We need $-1le3xle1$



    But if $xle0,$ the left hand side $le0$



    Now $3x=sin(pi/4-arcsin(2x))$



    $3sqrt2x=sqrt{1-(2x)^2}-2x$



    $sqrt{1-4x^2}=x(3sqrt2+2)$



    Square both sides






    share|cite





























      0














      Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
      $$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
      $$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
      $$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
      Square again, expand and simplify to get
      $$97 y^2-13 y+frac{1}{4}=0$$ which is simple.






      share|cite





















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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        There is a useful identity that we can use in this case:



        $arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$



        From here we can substitute:



        $arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$



        We are then left with:



        $2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$



        From here, you can solve for $x$.






        share|cite|improve this answer





















        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          1 hour ago


















        2














        There is a useful identity that we can use in this case:



        $arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$



        From here we can substitute:



        $arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$



        We are then left with:



        $2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$



        From here, you can solve for $x$.






        share|cite|improve this answer





















        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          1 hour ago
















        2












        2








        2






        There is a useful identity that we can use in this case:



        $arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$



        From here we can substitute:



        $arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$



        We are then left with:



        $2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$



        From here, you can solve for $x$.






        share|cite|improve this answer












        There is a useful identity that we can use in this case:



        $arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$



        From here we can substitute:



        $arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$



        We are then left with:



        $2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$



        From here, you can solve for $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        GnumbertesterGnumbertester

        1285




        1285












        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          1 hour ago




















        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          1 hour ago


















        When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
        – user5722540
        1 hour ago






        When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
        – user5722540
        1 hour ago













        2














        Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
        $$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
        $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
        But I don't know how you would solve this last equation.






        share|cite|improve this answer























        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          59 mins ago
















        2














        Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
        $$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
        $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
        But I don't know how you would solve this last equation.






        share|cite|improve this answer























        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          59 mins ago














        2












        2








        2






        Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
        $$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
        $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
        But I don't know how you would solve this last equation.






        share|cite|improve this answer














        Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
        $$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
        $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
        But I don't know how you would solve this last equation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        D.B.D.B.

        1,0748




        1,0748












        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          59 mins ago


















        • When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
          – user5722540
          59 mins ago
















        When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
        – user5722540
        59 mins ago




        When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
        – user5722540
        59 mins ago











        1














        I'm gonna derive the general function for $arcsin x$ then go from there.



        Recall that
        $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
        So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
        Letting $u=e^{iy}$, we have
        $$2ix=frac{u^2-1}{u}$$
        $$u^2-2ixu-1=0$$
        Use the quadratic formula to find that
        $$u=ix+sqrt{1-x^2}$$
        Thus
        $$e^{iy}=ix+sqrt{1-x^2}$$
        $$iy=lnbig[ix+sqrt{1-x^2}big]$$
        $$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
        So we look at your equation:
        $$arcsin 2x+arcsin3x=fracpi4$$
        $$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
        $$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
        Using the property $ln(ab)=ln a+ln b$ we see that
        $$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
        Taking $exp$ on both sides,
        $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
        Use the formula $e^{itheta}=costheta+isintheta$ to see that
        $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
        and at this point I used Wolfram|Alpha to see that
        $$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
        I will update my answer once I figure out how this result is found






        share|cite|improve this answer


























          1














          I'm gonna derive the general function for $arcsin x$ then go from there.



          Recall that
          $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
          So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
          Letting $u=e^{iy}$, we have
          $$2ix=frac{u^2-1}{u}$$
          $$u^2-2ixu-1=0$$
          Use the quadratic formula to find that
          $$u=ix+sqrt{1-x^2}$$
          Thus
          $$e^{iy}=ix+sqrt{1-x^2}$$
          $$iy=lnbig[ix+sqrt{1-x^2}big]$$
          $$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
          So we look at your equation:
          $$arcsin 2x+arcsin3x=fracpi4$$
          $$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
          $$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
          Using the property $ln(ab)=ln a+ln b$ we see that
          $$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
          Taking $exp$ on both sides,
          $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
          Use the formula $e^{itheta}=costheta+isintheta$ to see that
          $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
          and at this point I used Wolfram|Alpha to see that
          $$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
          I will update my answer once I figure out how this result is found






          share|cite|improve this answer
























            1












            1








            1






            I'm gonna derive the general function for $arcsin x$ then go from there.



            Recall that
            $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
            So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
            Letting $u=e^{iy}$, we have
            $$2ix=frac{u^2-1}{u}$$
            $$u^2-2ixu-1=0$$
            Use the quadratic formula to find that
            $$u=ix+sqrt{1-x^2}$$
            Thus
            $$e^{iy}=ix+sqrt{1-x^2}$$
            $$iy=lnbig[ix+sqrt{1-x^2}big]$$
            $$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
            So we look at your equation:
            $$arcsin 2x+arcsin3x=fracpi4$$
            $$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
            $$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
            Using the property $ln(ab)=ln a+ln b$ we see that
            $$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
            Taking $exp$ on both sides,
            $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
            Use the formula $e^{itheta}=costheta+isintheta$ to see that
            $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
            and at this point I used Wolfram|Alpha to see that
            $$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
            I will update my answer once I figure out how this result is found






            share|cite|improve this answer












            I'm gonna derive the general function for $arcsin x$ then go from there.



            Recall that
            $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
            So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
            Letting $u=e^{iy}$, we have
            $$2ix=frac{u^2-1}{u}$$
            $$u^2-2ixu-1=0$$
            Use the quadratic formula to find that
            $$u=ix+sqrt{1-x^2}$$
            Thus
            $$e^{iy}=ix+sqrt{1-x^2}$$
            $$iy=lnbig[ix+sqrt{1-x^2}big]$$
            $$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
            So we look at your equation:
            $$arcsin 2x+arcsin3x=fracpi4$$
            $$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
            $$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
            Using the property $ln(ab)=ln a+ln b$ we see that
            $$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
            Taking $exp$ on both sides,
            $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
            Use the formula $e^{itheta}=costheta+isintheta$ to see that
            $$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
            and at this point I used Wolfram|Alpha to see that
            $$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
            I will update my answer once I figure out how this result is found







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 21 mins ago









            clathratusclathratus

            3,320331




            3,320331























                0














                We need $-1le3xle1$



                But if $xle0,$ the left hand side $le0$



                Now $3x=sin(pi/4-arcsin(2x))$



                $3sqrt2x=sqrt{1-(2x)^2}-2x$



                $sqrt{1-4x^2}=x(3sqrt2+2)$



                Square both sides






                share|cite


























                  0














                  We need $-1le3xle1$



                  But if $xle0,$ the left hand side $le0$



                  Now $3x=sin(pi/4-arcsin(2x))$



                  $3sqrt2x=sqrt{1-(2x)^2}-2x$



                  $sqrt{1-4x^2}=x(3sqrt2+2)$



                  Square both sides






                  share|cite
























                    0












                    0








                    0






                    We need $-1le3xle1$



                    But if $xle0,$ the left hand side $le0$



                    Now $3x=sin(pi/4-arcsin(2x))$



                    $3sqrt2x=sqrt{1-(2x)^2}-2x$



                    $sqrt{1-4x^2}=x(3sqrt2+2)$



                    Square both sides






                    share|cite












                    We need $-1le3xle1$



                    But if $xle0,$ the left hand side $le0$



                    Now $3x=sin(pi/4-arcsin(2x))$



                    $3sqrt2x=sqrt{1-(2x)^2}-2x$



                    $sqrt{1-4x^2}=x(3sqrt2+2)$



                    Square both sides







                    share|cite












                    share|cite



                    share|cite










                    answered 5 mins ago









                    lab bhattacharjeelab bhattacharjee

                    223k15156274




                    223k15156274























                        0














                        Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
                        $$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
                        $$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
                        $$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
                        Square again, expand and simplify to get
                        $$97 y^2-13 y+frac{1}{4}=0$$ which is simple.






                        share|cite


























                          0














                          Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
                          $$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
                          $$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
                          $$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
                          Square again, expand and simplify to get
                          $$97 y^2-13 y+frac{1}{4}=0$$ which is simple.






                          share|cite
























                            0












                            0








                            0






                            Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
                            $$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
                            $$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
                            $$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
                            Square again, expand and simplify to get
                            $$97 y^2-13 y+frac{1}{4}=0$$ which is simple.






                            share|cite












                            Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
                            $$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
                            $$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
                            $$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
                            Square again, expand and simplify to get
                            $$97 y^2-13 y+frac{1}{4}=0$$ which is simple.







                            share|cite












                            share|cite



                            share|cite










                            answered 1 min ago









                            Claude LeiboviciClaude Leibovici

                            119k1157132




                            119k1157132






























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