# How do you solve cos2x + sinx = 0 over the interval 0 to 2pi?

Apr 5, 2016

Change $\cos \left(2 x\right)$ to $\sin \left(x\right)$ using the double angle formula and solve the resulting quadratic equation.

#### Explanation:

Double angle formula for cosine

$\cos \left(2 x\right) \equiv 1 - 2 {\sin}^{2} \left(x\right)$



Replace the $\cos \left(2 x\right)$ in the equation.

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

$= 0$



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

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



Therefore, either

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



$\sin \left(x\right) = 1$ corresponds to $x = {\sin}^{- 1} \left(1\right) = \frac{\pi}{2}$.



For $\sin \left(x\right) = - \frac{1}{2}$, the basic angle is ${\sin}^{- 1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$. Since $\sin \left(x\right)$ is negative when $x$ is in the third or forth quadrant,

$x = \pi + \frac{\pi}{6} = \frac{7 \pi}{6}$

or

$x = 2 \pi - \frac{\pi}{6} = \frac{13 \pi}{6}$



Putting together all the answers, $x = \frac{\pi}{2} , \frac{7 \pi}{6} \mathmr{and} \frac{13 \pi}{6}$.



Refer to the $x$-intercepts of the graph $y = \cos \left(2 x\right) + \sin \left(x\right)$ below.
graph{cos(2x)+sin(x) [-10, 10, -5, 5]}