# Solve the equation cos2theta+sintheta=0, if 0<=theta<=2pi?

Jan 7, 2018

$\theta = \frac{\pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$

#### Explanation:

cos2θ + sinθ = 0
$\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$
:. cos2θ + sinθ=1-2sin^2theta+sintheta=0
(rearranging to look more like a quadratic)
$- 2 {\sin}^{2} \theta + \sin \theta + 1 = 0$
$\textcolor{red}{- 1} \left(- 2 {\sin}^{2} \theta + \sin \theta + 1\right) = \textcolor{red}{- 1} \times 0$
$2 {\sin}^{2} \theta - \sin \theta - 1 = 0$
Let $x = \sin \theta$
$2 {x}^{2} - x - 1 = 0$
Factorising the quadratic yields;
$\left(2 x + 1\right) \left(x - 1\right) = 0$
By the null factor law;
$x = - \frac{1}{2}$ or $x = 1$
$\therefore \sin \theta = - \frac{1}{2}$ or $\sin \theta = 1$
Hence,
$\theta = \frac{\pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$ on the domain $\left[0 , 2 \pi\right]$

I hope that helps :)

Jan 7, 2018

$\theta = \frac{\pi}{2}$ or $\frac{7 \pi}{6}$ or $\frac{11 \pi}{6}$

#### Explanation:

We have

$\cos 2 \theta + \sin \theta = 0$, solutions in $0 \le \theta \le 2 \pi$

Substitute ${\cos}^{2} \theta = 1 - 2 {\sin}^{2} \theta$

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

Let $\sin \theta = u$

$1 - 2 {u}^{2} + u = 0$

$\iff 2 {u}^{2} - u - 1 = 0$

or $\left(2 u + 1\right) \left(u - 1\right) = 0$

$u = 1 , u = - \frac{1}{2}$

Substitute back $\sin \theta = 1$ or $- \frac{1}{2}$

and for $0 \le \theta \le 2 \pi$

$\theta = \frac{\pi}{2}$ or $\pi + \frac{\pi}{6} = \frac{7 \pi}{6}$ or $2 \pi - \frac{\pi}{6} = \frac{11 \pi}{6}$