СРОЧНО С1!!88 баллов даю!!!!

A) $16\cos^4x-24\cos^2x+9=0;\\ \cos x=t,\ \ \ \ -1\leq x\leq1;\\ 16t^4-24t^2+9=0;\\ (4t^2)^2-2\cdot4t^2\cdot3+3^2=0;\\ (4t^2-3)=0;\\ 4t^2-3=0;\\ 4t^2=3;\\ t^2=\frac{3}{4};\\ t=\pm\frac{\sqrt3}{2};\\$
$\cos x=\pm\frac{\sqrt3}{2};\\ \cos x=\frac{\sqrt3}{2};\\ x=\pm\arccos\frac{\sqrt3}{2}+2\pi n, n\in Z;\\ x=\pm\frac{\pi}{6}+2\pi n, n\in Z;\\ \cos x=-\frac{\sqrt3}{2};\\ x=\pm\arccos(-\frac{\sqrt3}{2})+2\pi k, k \in Z;\\ x=\pm\frac{5\pi}{6}+2\pi k;\\ \left[ {{x=\pm\frac{\pi}{6}+2\pi n,} \atop {x=\pi\pm\frac{\pi}{6}+2\pi k}} \right. \ \ \ n,k\in Z$

б)
$x\in [2\pi,3\pi];\\ 2\pi+\frac{\pi}{6}=\frac{12\pi+\pi}{6}=\frac{13\pi}{6};\\ 2\pi+\pi-\frac{\pi}{6}=3\pi-\frac{\pi}{6}=\frac{18\pi-\pi}{6}=\frac{17\pi}{6};\\ \\ \\ \\ x=\frac{13\pi}{6};\\ x=\frac{17\pi}{6};\\$