Question

Solve 3cos^2θ + 5cosθ + 2 = 0 for 0° ≤ θ < 180°.

ans:Ø,{109°},{132°},{180°}

Answer 1

3cos^2θ + 5cosθ + 2
3(2cos^2(θ)-1)+5cosθ+2=0
6cos^2(θ)+5cosθ-1=0
(6cosθ-1)(cosθ+1)=0
cosθ=1/6 or cosθ=-1
since cosθ<180 we can neglect cosθ=-1
then right the general solution for cosθ=1/6..
You will get the answers :)