Answer:
[tex]\frac{7\pi }{6} ; \ \frac{3\pi }{2} ; \ \frac{11\pi }{6}.[/tex]
Step-by-step explanation:
if 1=sin²Q+cos²Q, then cos²Q=1-sin²Q, then
2sin²Q+3sinQ+1=0;
[tex]\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*\frac{7\pi }{6}+\pi n \end{array}[/tex]
on the interval Q=3π/2 (from the 1st equation) and Q=7π/6; 11π/6 (from the 2d equation).
