Question

1)cos3b(бета)sin(3(pi++1,5sin6b 2)(tg2(в квадрате)(x+pi/4)-1)(cos2x(в )=0

Answer 1

$1.\; \; cos3 \beta \cdot sin(3(\pi + \beta ))-2+1,5sin6 \beta =\\\\=\frac{1}{2}(sin(6 \beta +3\pi )+sin(3\pi +3 \beta -3 \beta ))-2+1,5sin6 \beta =\\\\=-\frac{1}{2}sin6 \beta +\frac{1}{2}sin3\pi -2+1,5sin6 \beta =\\\\=sin6 \beta -2\; \; (sin3\pi =0)\\\\2.\; \; (tg^2(x+\frac{\pi}{4})-1)(cos^2x-4)=0\\\\a)\; cos^2x-4=0,\; \to \; \; cos^2x=4,\; cosx=\pm 2\; net\; reshenij\; (|cosx| \leq 1)\\\\b)tg^2(x+\frac{\pi}{4})=1,tg(x+\frac{\pi}{4})=\pm 1\\\\x+\frac{\pi}{4}=\frac{\pi}{4}+\pi n,\; n\in Z$

$x=\pi n\\\\x+\frac{\pi}{4}=-\frac{\pi}{4}+\pi k,\; k\in Z\\\\x=-\frac{\pi}{2}+\pi k\\\\otvet:\; x=\pi n,n\in Z;\; \; x=-\frac{\pi}{2}+\pi k,k\in Z$