How would you solve for x in this case using the trig identities? 
Solve for all values of $x$:
$$\cos 2x = 2\sin x$$
$$1 - 2\sin^2x - 2\sin x = 0$$
$$-2\sin^2x - 2\sin x + 1 = 0$$
$$\sin^2x - 2\sin x - 2 = 0$$

How would you factor this above to solve for $x$?
 A: $$\cos(2x)=2\sin(x)\Longleftrightarrow$$
$$\cos(2x)-2\sin(x)=0\Longleftrightarrow$$
$$1-2\sin(x)-2\sin^2(x)=0\Longleftrightarrow$$
$$-\frac{1}{2}+\sin(x)+\sin^2(x)=0\Longleftrightarrow$$
$$\sin(x)+\sin^2(x)=\frac{1}{2}\Longleftrightarrow$$
$$\frac{1}{4}+\sin(x)+\sin^2(x)=\frac{3}{4}\Longleftrightarrow$$
$$\left(\frac{1}{2}+\sin(x)\right)^2=\frac{3}{4}\Longleftrightarrow$$
$$\frac{1}{2}+\sin(x)=\pm\frac{\sqrt{3}}{2}\Longleftrightarrow$$
$$\sin(x)=\pm\frac{\sqrt{3}}{2}-\frac{1}{2}\Longleftrightarrow$$
$$x=\begin{cases}
\ \pi+\arcsin\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)+2\pi n_1 \\
\ \pi+\arcsin\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)+2\pi n_2 \\
\ 2\pi n_3-\arcsin\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right) \\
\ 2\pi n_4-\arcsin\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)
\end{cases}$$
With $n_1,n_2,n_,n_4\in\mathbb{Z}$
A: Hint: $\cos 2x = 1 - 2\sin^2 x$
A: HINT
Solve quadratic equation for $\sin x$ :
$$ 1- 2 s^2 - 2 s = 0 $$
A: $$\cos 2x = 2\sin x$$
or, $$1-2\sin^2 x = 2\sin x$$
or, $$2\sin^2 x + 2\sin x - 1 = 0$$
So $$\sin x = \frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}=\omega, \omega^2$$
A: First there needs to be a little algebra and trigonometry,
$$\cos(2x)=2\sin(x)$$
$$1-2\sin^2(x)=2\sin(x)$$
$$1=2\sin^2(x)-2\sin(x)$$
$$2\sin^2(x)-2\sin(x)-1=0$$
$$\sin x = \frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}$$
Let $z=\sin(x)$. Then,
$$2\sin^2(x)-2\sin(x)-1=0\implies2z^2-2z-1=0$$
Now use quadratic formula,
$$z=\frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}=z$$ 
Since $z=\sin(x)$, then
$$x=\sin^{-1}(z)$$
$$x=\sin^{-1}(\frac{-1\pm\sqrt{3}}{2})$$
$$x=\sin^{-1}(\frac{-1+\sqrt{3}}{2}), \ x=\sin^{-1}(\frac{-1-\sqrt{3}}{2})$$
The rest is trivial: just calculate!
