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=- (1)/(2) \\ \theta= ( 2\pi )/(3) +2 \pi k~~~k \in Z \\ \\ 0 \leq \theta \ \textless \ 360^(а) \\ 0 \leq \theta \ \textless \ 2 \pi \\ \theta = (2 \pi )/(3) \\ \theta = (2 \pi )/(3) , (4 \pi )/(3)`