Question

Solve Sine inverse x +sine inverse 2x = π/3

Answer 1

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

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

$\left(8x^4-5x^2+\dfrac34\right)^2=16x^4(1-4x^2)(1-x^2)$
$64x^8-80x^6+37x^4-\dfrac{15}2x^2+\dfrac9{16}=64x^8-80x^6+16x^4$
$21x^4-\dfrac{15}2x^2+\dfrac9{16}=0$
$112x^4-40x^2+3=0$
$(4x^2-1)(28x^2-3)=0$
$(2x-1)(2x+1)(\sqrt{28}x-\sqrt3)(\sqrt{28}x+\sqrt3)=0$

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

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.