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Mrblunt5613

08.06.2020

Mathematics

$\arcsin x+\arcsin2x=\dfrac\pi3$
$\sin\left(\arcsin x+\arcsin2x\right)=\sin\dfrac\pi3$
$\sin(\arcsin x)\cos(\arcsin2x)+\sin(\arcsin2x)\cos(\arcsin x)=\dfrac{\sqrt3}2$

Note that $\arcsin x$ is defined for $|x|\le1$ , and $\arcsin2x$ is defined for $|x|\le\dfrac12$ , where the latter will be the "total" domain. Under this condition, you have $\sin(\arcsin x)=x$ and $\sin(\arcsin2x)=2x$ . The cosine terms can be found with Pythagoras' theorem.

So provided that $|x|\le\dfrac12$ , it follows that the above reduces to

$x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt3}2$

Squaring both sides gives

$x^2(1-4x^2)+4x^2\sqrt{(1-4x^2)(1-x^2)}+4x^2(1-x^2)=\dfrac34$
$5x^2-8x^4+4x^2\sqrt{(1-4x^2)(1-x^2)}=\dfrac34$
$8x^4-5x^2+\dfrac34=4x^2\sqrt{(1-4x^2)(1-x^2)}$

Squaring both sides again gives

$\left(8x^4-5x^2+\dfrac34\right)^2=16x^4(1-4x^2)(1-x^2)$
$64x^8-80x^6+37x^4-\dfrac{15}2x^2+\dfrac9{16}=64x^8-80x^6+16x^4$
$21x^4-\dfrac{15}2x^2+\dfrac9{16}=0$
112x^4-40x^2+3=0

$(2x-1)(2x+1)(\sqrt{28}x-\sqrt3)(\sqrt{28}x+\sqrt3)=0$

$\implies x=\pm\dfrac12,\pm\sqrt{\dfrac3{28}}$

Three of these solutions are extraneous, however.

When $x=-\dfrac12$ , we have $\arcsin x+\arcsin2x=-\dfrac{2\pi}3$ .

When $x=\dfrac12$ , we have $\arcsin x+\arcsin2x=\dfrac{2\pi}3$ .

When $x=-\sqrt{\dfrac3{28}}$ , we have $\arcsin x+\arcsin2x=-\dfrac\pi3$ .

Finally, when $x=\sqrt{\dfrac3{28}}$ , we have $\arcsin x+\arcsin2x=\dfrac\pi3$ , so this is our only solution.

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