How do you solve sin2θ=-cosθ ?

Apr 27, 2018

$\theta = \left\{\left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}\right\} \cup \left\{k \pi - {\left(- 1\right)}^{k} \cdot \frac{\pi}{6} , k \in \mathbb{Z}\right\}$

Explanation:

We know that,

color(red)(sinx=sinalpha=>x=kpi+(-1)^k*alpha,kinZZ

Here,

$\sin 2 \theta = - \cos \theta$

$\implies \sin 2 \theta + \cos \theta = 0$

$\implies 2 \sin \theta \cos \theta + \cos \theta = 0$

$\implies \cos \theta \left(2 \sin \theta + 1\right) = 0$

$\implies \cos \theta = 0 \mathmr{and} 2 \sin \theta + 1 = 0$

$= \cos \theta = 0 \mathmr{and} \sin \theta = - \frac{1}{2}$

(i)color(blue)(costheta=0=>theta=(2k+1)pi/2,kinZZ

$\left(i i\right) \sin \theta = - \frac{1}{2} \implies \sin \theta = \sin \left(- \frac{\pi}{6}\right)$

$\implies \theta = k \pi + {\left(- 1\right)}^{k} \left(- \frac{\pi}{6}\right) , k \in \mathbb{Z}$

=>color(blue)(theta=kpi-(-1)^k(pi/6),kinZZ

Hence,

$\theta = \left\{\left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}\right\} \cup \left\{k \pi - {\left(- 1\right)}^{k} \cdot \frac{\pi}{6} , k \in \mathbb{Z}\right\}$

Apr 27, 2018

Within the interval of $0 < \theta < 2 \pi$

$\theta = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$

Explanation:

.

We use the double-angle formula:

$\sin 2 \theta = - \cos \theta$

$2 \sin \theta \cos \theta = - \cos \theta$

We move all terms to one side.

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

We factor out $\cos \theta$

$\cos \theta \left(2 \sin \theta + 1\right) = 0$

Then, we set each piece equal to $0$ and solve for $\theta$.

$\cos \theta = 0 , \therefore \theta = \frac{\pi}{2} , \frac{3 \pi}{2}$

$2 \sin \theta + 1 = 0$

$\sin \theta = - \frac{1}{2} , \therefore \theta = \frac{7 \pi}{6} , \frac{11 \pi}{6}$