# Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ [closed]

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?

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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?

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$1 - 2\sin^2 x = 2\cos^2 x - 1\\ 2\cos^2 x - 1 = \pm \cos x\\ 2\cos^2 x \pm\cos x - 1 = 0$

$\cos x = \pm \frac 14 \pm \frac 34\\ \cos x = \pm \frac 12, \pm 1$

HINT:

Solve each of the following equations \begin{align}1-2\sin^2x&=\cos x\\1-2\sin^2x&=-\cos x\end{align}

• only two equations are enough, right? Apr 25, 2018 at 20:30

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}

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}

• Actually $\sin3x=0\implies 3x=m\pi$ Apr 26, 2018 at 3:49