А) 2cos(pi/4+x)=корень из 2 б) -2sin(2x+pi/3)-1=0

а) 2 cos (π/4 +x) = √2 cos (π/4 +x) = √2/2 π/4 + x = π/4 + 2πm, m ∈ Z - 1 корень π/4 + x = - π/4 + 2πn, n ∈ Z - 2 корень π/4 + x - π/4 = 2πm, m ∈ Z  π/4 + x + π/4 = 2πn, n ∈ Z  x = 2πm, m ∈ Z x = 2πn - π/2, n ∈ Z Ответ: -π/2 + 2πn, n∈Z; 2πm, m ∈ Z

[latex]a)\; 2cos(\frac{\pi}{4}+x)=\sqrt 2\\cos(\frac{\pi}{4}+x)=\frac{\sqrt2}{2}\\\frac{\pi}{4}+x=\pm \frac{\pi}{4}+2\pi n\\\\x_1=\frac{\pi}{4}-\frac{\pi}{4}+2\pi n\\x_1=2\pi n, \; n\in Z;\\\\x_2=-\frac{\pi}{4}-\frac{\pi}{4}+2\pi n\\x_2=-\frac{\pi}{2}+2\pi n, \; n\in Z;\\\\ b)\;-2sin(2x+\frac{\pi}{3})-1=0\\sin(2x+\frac{\pi}{3})=-\frac{1}{2}\\2x+\frac{\pi}{3}=(-1)^{n+1}\frac{\pi}{3}+\pi n\\2x=(-1)^{n+1}\frac{\pi}{3}-\frac{\pi}{3}+\pi n\\x=\frac{1}{2}*(-1)^{n+1}\frac{\pi}{3}-\frac{\pi}{6}+\frac{\pi n}{2},\;n\in Z.[/latex]