Confusing trigonometric double angle identity?

Question

How can I arrange $\sin^22x+3\sin^2x=3$ so that $\sin x=a/b$ or something similar.

Answer 1

$$\sin^2 2x+3\sin^2x=3$$ $$ \Rightarrow (2\sin x\cos x)^2 + 3\sin^2x - 3 = 0$$ $$ \Rightarrow 4\sin^2x\cos^2x + 3\sin^2x - 3 = 0$$ $$ \Rightarrow 4\sin^2x(1-\sin^2x) + 3\sin^2x - 3 = 0$$ $$ \Rightarrow 4\sin^2x-4\sin^4x + 3\sin^2x - 3 = 0$$ $$ \Rightarrow -4\sin^4x + 7\sin^2x - 3 = 0$$ Now make the substitution $\sin^2x = u$ $$ \Rightarrow -4u^2 + 7u - 3 = 0$$ $$ \Rightarrow -(u-1)(4u-3)=0$$ $$ \Rightarrow -(u-1)(4u-3)=0$$ $$ \Rightarrow u=1, 4u-3=0$$ $$ \Rightarrow \sin^2x=1$$ $$ \Rightarrow \sin x=1,\ \sin x=-1$$ $$ \Rightarrow x=\frac{\pi}{2}+2\pi n_1, \,\,x = \pi n_2 - \frac{\pi}{2}\,\,**$$ $$4u-3=0 \Rightarrow 4u=3$$ $$\Rightarrow u = \frac{3}{4} = \sin^2x$$ $$\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}$$ $$\Rightarrow \sin x = \frac{\sqrt{3}}{2}$$ $$\Rightarrow x = \frac{2\pi}{3}+2\pi n_3, \,\, x = \frac{\pi}{3}+2\pi n_4 \,\,**$$ $$\sin x = -\frac{\sqrt{3}}{2}$$ $$\Rightarrow x = \frac{4\pi}{3}+2\pi n_5, \,\, x = \frac{5\pi}{3}+2\pi n_6\,\,**$$

Answer 2

Hint:

You can use $\sin^2 2x+3\sin^2x=3\iff1-\cos^22x+3\cdot\frac{1}{2}(1-\cos 2x)=3$

$\hspace{2.3 in}\iff2\cos^2 2x+3\cos2x+1=0$