Решение:

$\sin2x+1=\sin x+\cos x;$ $2s\mathrm{in}x\cos x+\cos^2x+\sin^2x=\sin x+\cos x;$$(\sin x+\cos x)^2=\sin x+\cos x;$$(\sin x+\cos x)(\sin x+\cos x-1)=0;$

$\begin{array}{l}\left\{\begin{array}{l}\sin x+\cos x=0\\\sin x+\cos x=1\end{array}\right.\\\left\{\begin{array}{l}\frac{\sqrt2}2\sin x+\frac{\sqrt2}2\cos x=0\\\frac{\sqrt2}2\sin x+\frac{\sqrt2}2\cos x=\frac{\sqrt2}2\end{array}\right.\\\left\{\begin{array}{l}\sin(x+\frac{\mathrm\pi}4)=0\\\sin(x+\frac{\mathrm\pi}4)=\frac{\sqrt2}2\end{array}\right.\\\left\{\begin{array}{l}x=-\frac{\mathrm\pi}4+\mathrm{πn},\;\mathrm n\in\mathbb{Z}\\x=-\frac{\mathrm\pi}4+(-1)^k\frac{\mathrm\pi}4+\mathrm{πk},\;\mathrm n\in\mathbb{Z}\end{array}\right.\\\end{array}$

б) Найдем все корни уравнения, принадлежащие промежутку $\lbrack\frac{\mathrm\pi}2;2\mathrm\pi)$

$\begin{array}{l}n=1,\;x=\frac{3\mathrm\pi}4\\n=2,\;x=\frac{7\mathrm\pi}4\\k=1,\;x=\frac{\mathrm\pi}2\end{array}$

Ответ: а) $-\frac\pi4+\pi\kappa,\;-\frac{\mathrm\pi}4+(-1)^k\frac{\mathrm\pi}4+\mathrm{πk},\;\mathrm n\in\mathbb{Z};$

б) $\frac\pi2;\;\frac34\pi;\;\frac74\pi$