11.1. For all possible values of the parameter $a$, solve the equation

$\log _{2} \sqrt{\cos x}=\log _{4}(a \cdot \sin x)$
Answer. When $a=0$ the equation makes no sense;

when $a>0 \quad x=\operatorname{arcctg} a+2 \pi n, n \in \mathbb{Z}$;

when $a<0 \quad x=\pi+\operatorname{arcctg} a+2 \pi m, m \in \mathbb{Z}$. On the domain of definition, we have the trigonometric equation $\cos x=a \sin x \Leftrightarrow$ $\operatorname{ctg} x=a \Leftrightarrow x=\operatorname{arctg} a+\pi n, n \in \mathrm{Z}$ and, taking into account the domain of definition, we will have: 1) when $a>0 \quad x=\operatorname{arcctg} a+2 \pi n, n \in Z$ (since otherwise $x$ lies in the third quadrant, and then $a \sin x<0$); 2) when $a<0 \quad x=\pi+\operatorname{arcctg} a+2 \pi m, m \in \mathrm{Z}$ (since otherwise $x$ lies in the second quadrant, and then $a \sin x<0$).