Буду благодарен за помощь. решить уравнение. 2cos(x- \frac{ \pi }{4} )=1 2sin(3x-\frac{...

Решите задачу:

$2cos(x- \frac{ \pi }{4} )=1\\cos(x-\frac{\pi}{4})=\frac{1}{2}\\x-\frac{\pi}{4}=\pm \frac{\pi}{3}+2\pi n\\x_1=\frac{7\pi}{12}+2\pi n, \; n\in Z;\\x_2=-\frac{\pi}{12}+2\pi n, \; n\in Z;\\\\ 2sin(3x-\frac{ \pi }{8}) =1\\sin(3x-\frac{\pi}{8})=\frac{1}{2}\\3x-\frac{\pi}{8}=(-1)^n\frac{\pi}{6}+\pi n\\3x=(-1)^n\frac{\pi}{6}+\frac{\pi}{8}+\pi n\\x=\frac{1}{3}*(-1)^n\frac{\pi}{6}+\frac{\pi}{24}+\frac{\pi n}{3}, \; n\in Z.$

$(1-ctg 45^o+cos60^o)* \frac{1}{2} -sin90^o*tg30^o=\\(1-1+\frac{1}{2})*\frac{1}{2}-1*\frac{\sqrt3}{3}=\frac{1}{4}-\frac{\sqrt3}{3}=\frac{3-4\sqrt3}{12}.$