[tex]\arcsin x+\arcsin2x=\dfrac\pi3[/tex]
[tex]\sin\left(\arcsin x+\arcsin2x\right)=\sin\dfrac\pi3[/tex]
[tex]\sin(\arcsin x)\cos(\arcsin2x)+\sin(\arcsin2x)\cos(\arcsin x)=\dfrac{\sqrt3}2[/tex]
Note that [tex]\arcsin x[/tex] is defined for [tex]|x|\le1[/tex], and [tex]\arcsin2x[/tex] is defined for [tex]|x|\le\dfrac12[/tex], where the latter will be the "total" domain. Under this condition, you have [tex]\sin(\arcsin x)=x[/tex] and [tex]\sin(\arcsin2x)=2x[/tex]. The cosine terms can be found with Pythagoras' theorem.
So provided that [tex]|x|\le\dfrac12[/tex], it follows that the above reduces to
[tex]x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt3}2[/tex]
Squaring both sides gives
[tex]x^2(1-4x^2)+4x^2\sqrt{(1-4x^2)(1-x^2)}+4x^2(1-x^2)=\dfrac34[/tex]
[tex]5x^2-8x^4+4x^2\sqrt{(1-4x^2)(1-x^2)}=\dfrac34[/tex]
[tex]8x^4-5x^2+\dfrac34=4x^2\sqrt{(1-4x^2)(1-x^2)}[/tex]
Squaring both sides again gives
[tex]\left(8x^4-5x^2+\dfrac34\right)^2=16x^4(1-4x^2)(1-x^2)[/tex]
[tex]64x^8-80x^6+37x^4-\dfrac{15}2x^2+\dfrac9{16}=64x^8-80x^6+16x^4[/tex]
[tex]21x^4-\dfrac{15}2x^2+\dfrac9{16}=0[/tex]
[tex]112x^4-40x^2+3=0[/tex]
[tex](4x^2-1)(28x^2-3)=0[/tex]
[tex](2x-1)(2x+1)(\sqrt{28}x-\sqrt3)(\sqrt{28}x+\sqrt3)=0[/tex]
[tex]\implies x=\pm\dfrac12,\pm\sqrt{\dfrac3{28}}[/tex]
Three of these solutions are extraneous, however.
When [tex]x=-\dfrac12[/tex], we have [tex]\arcsin x+\arcsin2x=-\dfrac{2\pi}3[/tex].
When [tex]x=\dfrac12[/tex], we have [tex]\arcsin x+\arcsin2x=\dfrac{2\pi}3[/tex].
When [tex]x=-\sqrt{\dfrac3{28}}[/tex], we have [tex]\arcsin x+\arcsin2x=-\dfrac\pi3[/tex].
Finally, when [tex]x=\sqrt{\dfrac3{28}}[/tex], we have [tex]\arcsin x+\arcsin2x=\dfrac\pi3[/tex], so this is our only solution.
