4 sin^2 (x/2) + 8 cos (x/2) -7 = 0

Решите задачу:

$4sin^2\frac{x}{2}+8cos\frac{x}{2}-7=0\\\\4(1-cos^2\frac{x}{2})+8cos\frac{x}{2}-7=0\\\\t=cos\frac{x}{2}\; ,\; \; -4t^2+8t-3=0\; ,\; \; 4t^2-8t+3=0\; ,\; -1 \leq t \leq 1\\\\D/4=16-12=4\\\\t_1=\frac{4+2}{4}=\frac{3}{2}\; \; ne\; podxodit\\\\t_2= \frac{4-2}{4} =\frac{1}{2}\; \; \Rightarrow \; \; cos\frac{x}{2}=\frac{1}{2}\\\\\frac{x}{2}=\pm arccos\frac{1}{2}+2\pi n\; ,\; n\in Z\\\\\frac{x}{2}=\pm \frac{\pi}{3}+2\pi n\; ,\; n\in Z\\\\x=\pm \frac{2\pi }{3}+4\pi n\; ,\; n\in Z$