Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ I have difficulty in solving this equation:
$$|1-2\sin^2 x|=\lvert\cos x\rvert.$$
What are the indications to solve the exercise correctly?
 A: This is the same as saying: $$|\cos(2x)|=|\cos(x)|.$$ So either $\cos(2x)=\cos(x)$ and/or $\cos(2x)=-\cos(x)$. We already see $x=0$ in both equations and $x=\pm\pi$ in the second. Do you know more?
A: $1 - 2\sin^2 x = 2\cos^2 x - 1\\
2\cos^2 x  - 1 = \pm \cos x\\
2\cos^2 x  \pm\cos x - 1 = 0$
Apply the quadratic formula:
$\cos x = \pm \frac 14 \pm \frac 34\\
\cos x = \pm \frac 12, \pm 1 $
A: HINT: 
Solve each of the following equations $$\begin{align}1-2\sin^2x&=\cos x\\1-2\sin^2x&=-\cos x\end{align}$$
A: Using trigonometric equations 
$$ \begin{align} & |1-2\sin^2x| = |\cos(x)| \\  \iff & |\cos(2x)| = |\cos(x)|  \\
 \iff &\cos(2x) = - \cos(x) \mathrm{\ or\ } \cos(2x) =  \cos(x) \\
 \iff & \cos(2x) =  \cos(\pi+x) \mathrm{\ or\ } \cos(2x) =  \cos(x) \\
 \iff & 2x =  x +k\pi \mathrm{\ or\ }  2x =  -x +k\pi, \ k\in \mathbb Z  \\
 \iff & x = k\pi \mathrm{\ or\ } x = k\frac  \pi 3, \ k\in \mathbb Z
\end{align}$$
A: Since both sides are absolute values, you can square both sides:
$$\begin{align}
1-4\sin^2x+4\sin^4x&=\cos^2x \\
0&=-3\sin^2x+4\sin^4x \\
&=\sin^2x\left(4\sin^2x-3\right) \\
\sin x&=0,\pm\frac{\sqrt3}2
\end{align}$$

Following a comment by @labbhattacharjee, we note that $\sin(3x)=3\sin(x)-4\sin^3x$, so
$$\begin{align}
0&=-\sin(3x)\sin x \\
x&=\frac{\pi}3k  , \quad k\in\mathbb Z.
\end{align}$$
