Question

1)cos^2(7п+x)=1/2
2)sin^2(4,5п-x)=3/4

Answer 1

$\displaystyle\bf\\1) \ \ Cos^{2}\Big(7\pi +x\Big)=\frac{1}{2}\\\\\\\frac{1+Cos\Big(14\pi +2x\Big)}{2}=\frac{1}{2} \ |\cdot 2 \\\\\\1+Cos\Big(14\pi +2x\Big)=1\\\\\\1+Cos\Big(2\pi \cdot 7+2x\Big)=1\\\\\\1+Cos2x=1\\\\\\Cos2x=0\\\\\\2x=\frac{\pi }{2} +\pi n.n\in Z\\\\\\\boxed{x=\frac{\pi }{4} +\frac{\pi n}{2} ,n\in Z}$

$\displaystyle\bf\\2) \ \ Sin^{2} \Big(4,5\pi -x\Big)=\frac{3}{4}\\\\\\\frac{1-Cos\Big(9\pi -2x\Big)}{2}=\frac{3}{4} \ |\cdot 2\\\\\\1-Cos\Big(9\pi -2x\Big)=\frac{3}{2} \\\\\\1-Cos\Big[4\pi \cdot 2+(\pi -2x)\Big]=\frac{3}{2}\\\\\\1-Cos\Big(\pi -2x\Big)=\frac{3}{2} \\\\\\1+Cos2x=\frac{3}{2}\\\\\\Cos2x=\frac{1}{2}\\\\\\2x=\pm \ arcCos\frac{1}{2} +2\pi n,n\in Z\\\\\\2x=\pm \ \frac{\pi }{3} +2\pi n,n\in Z\\\\\\\boxed{x=\pm \ \frac{\pi }{6} +\pi n,n\in Z}$