# How do you solve cos 2x + sin x=0?

Apr 14, 2015

$x = {\sin}^{-} 1 \left(- \frac{1}{2}\right) , x = {\sin}^{-} 1 \left(1\right)$

Solution

$\cos 2 x + \sin x = 0$

As

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

So

${\cos}^{2} x - {\sin}^{2} x + \sin x = 0$

$1 - {\sin}^{2} x - {\sin}^{2} x + \sin x = 0$

$1 - 2 {\sin}^{2} x + \sin x = 0$

$- 2 {\sin}^{2} x + \sin x + 1 = 0$

$2 {\sin}^{2} x - \sin x - 1 = 0$

$2 {\sin}^{2} x - 2 \sin x + \sin x - 1 = 0$

$2 \sin x \left(\sin x - 1\right) + 1 \left(\sin x - 1\right)$

$\left(2 \sin x + 1\right) \left(\sin x - 1\right) = 0$

$\left(2 \sin x + 1\right) = 0 , \left(\sin x - 1\right) = 0$

$2 \sin x = - 1 , \sin x = 1$

$\sin x = - \frac{1}{2} , \sin x = 1$

$x = {\sin}^{-} 1 \left(- \frac{1}{2}\right) , x = {\sin}^{-} 1 \left(1\right)$

Apr 15, 2015

Replace $\cos 2 x = 1 - 2 {\sin}^{2} x$:

$f \left(x\right) = \cos 2 x + \sin x = 1 - 2 {\sin}^{2} x + \sin x = 0$

Call $\sin x = t$. This is a quadratic equation in $t$:

$f \left(t\right) = - 2 {t}^{2} + t + 1 = 0$

Solve this quadratic equation. Because $a + b + c = 0$, one real root is ${t}_{1} = 1$ and the other is ${t}_{2} = - \frac{1}{2}$

Next, solve the basic trig equation:

$t 1 = \sin x = 1 \to x = \frac{\pi}{2}$

Solve:

$t 2 = \sin x = - \frac{1}{2} \to x = \frac{7 \pi}{6} \mathmr{and} x = \frac{11 \pi}{6}$

Answers within period (0, 2pi): pi/2; (7pi)/6; and (11pi)/6

$x = \frac{\pi}{2} + k \cdot 2 \pi$
$x = 7 \frac{\pi}{6} + k \cdot 2 \pi$
$x = 11 \frac{\pi}{6} + k \cdot \pi$
$x = \frac{\pi}{2}$ --> $\cos 2 x = \cos \pi = - 1$; $\sin x = \sin \frac{\pi}{2} = 1$ --> f(x) = -1 + 1 = 0. OK
$x = \frac{7 \pi}{6}$ --> $\cos 2 x = \cos \left(\frac{14 \pi}{6}\right) = \cos \left(\frac{2 \pi}{6}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$; $\sin \left(\frac{7 \pi}{6}\right) = - \frac{1}{2}$ --> $f \left(x\right) = \frac{1}{2} - \frac{1}{2} = 0.$ OK
$x = \left(\frac{11 \pi}{6}\right)$ -> $\cos \left(\frac{22 \pi}{6}\right) = \cos \left(\frac{10 \pi}{6}\right) = \cos \frac{\left(5 \pi\right)}{3} = \frac{1}{2}$; $\sin \left(\frac{11 \pi}{6}\right) = - \frac{1}{2}$ --> $f \left(x\right) = \frac{1}{2} - \frac{1}{2} = 0$. OK.