Question

Solve the question on the interval [0, 2π)
sin^2 θ - cos^2 θ = -3 sin θ - 2

Answer 1

`(7\pi )/(6) ; \ (3\pi )/(2) ; \ (11\pi )/(6).`

Explanation:

if 1=sin²Q+cos²Q, then cos²Q=1-sin²Q, then

2sin²Q+3sinQ+1=0;

`\left[\begin{array}{ccc}sinQ=-1\\sinQ=-0.5\end{array} \ => \ \left[\begin{array}{ccc}Q=3\pi/2+2\pi *n\\Q=(-1)^n*(7\pi )/(6)+\pi n \end{array}`

on the interval Q=3π/2 (from the 1st equation) and Q=7π/6; 11π/6 (from the 2d equation).