
Problem 10b.1. Solve the equation:

$$
3^{\log _{3}(\cos x+\sin x)+\frac{1}{2}}-2^{\log _{2}(\cos x-\sin x)}=\sqrt{2}
$$


Solution: Note that the admissible values are those $x$ for which $\cos x+\sin x>0, \cos x-\sin x>0$. After simple calculations the equation becomes

$$
(\sqrt{3}-1) \cos x+(\sqrt{3}+1) \sin x=\sqrt{2}
$$

This equation is equivalent to

$$
\frac{\sqrt{6}-\sqrt{2}}{4} \cos x+\frac{\sqrt{6}+\sqrt{2}}{4} \sin x=\frac{1}{2}
$$

Since $\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$ we have

$$
\sin \left(x+15^{\circ}\right)=\frac{1}{2}
$$

From the roots of the later equation only $x=15^{\circ} \pm k 360^{\circ}$ are admissible. Therefore the roots are $x=15^{\circ} \pm k 360^{\circ}$.
