Помогите решить 1 номер, пожалуйста!!((

Решите задачу:

$y=\frac{x}{\sqrt2}-\cos x, [-\pi;\pi]\\ y'=\frac{1}{\sqrt2}+\sin x\\ \sin x=-\frac{\sqrt2}{2}\\ x=\frac{5\pi}{4}+2\pi*n, n\in Z\\ x=\frac{7\pi}{4}+2\pi*n, n\in Z\\ x\in [-\pi;\pi]:\\ x_1=-\frac{\pi}{4},\;\;\;\;\; x_2=-\frac{3\pi}{4}\\ y(-\pi)=\frac{-\pi}{\sqrt2}-\cos (-\pi)=\frac{-\pi}{\sqrt2}+1\approx{-1.22} \\ y(\pi)=\frac{\pi}{\sqrt2}-\cos (\pi)=\frac{\pi}{\sqrt2}+1\approx{3.22}\\ y(-\frac{\pi}{4})=\frac{-\pi}{4\sqrt2}-\cos (-\frac{\pi}{4})=\frac{-\pi}{4\sqrt2}-\frac{\sqrt2}{2}\approx{-1.82}\\ y(-\frac{3\pi}{4})=\frac{-3\pi}{4\sqrt2}-\cos (-\frac{3\pi}{4})=\frac{-3\pi}{4\sqrt2}+\frac{\sqrt2}{2}\approx{-0.96}\\ y_{max}=\frac{\pi}{\sqrt2}+1,\; \; \; \; \; y_{min}=\frac{-\pi}{4\sqrt2}-\frac{\sqrt2}{2}$