Question

Solve for 2cos^2theta + 13costheta = -6. theta is >= 0 and < 360 degrees

Answer 1

$\\ 2cos^2 \theta+13cos \theta=-6 \\ 2cos^2 \theta+13cos \theta+6=0 \\ 2cos^2 \theta+12cos \theta+1cos \theta+6=0 \\ 2cos \theta(cos \theta+6)+1(cos \theta+6)=0 \\ (2cos\theta +1)(cos\theta+6)=0$

$cos\theta+6=0 \\ cos\theta=-6 \\ \\ cos(x)=|a| \leq 1 \\ |-6| \geq 1$


$2cos \theta+1=0 \\ 2cos \theta=-1 \\ cos \theta=- \frac{1}{2} \\ \theta= \frac{ 2\pi }{3} +2 \pi k~~~k \in Z \\ \\ 0 \leq \theta \ \textless \ 360^{а} \\ 0 \leq \theta \ \textless \ 2 \pi \\ \theta = \frac{2 \pi }{3} \\ \theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$