Помогите ,пожалуйста sin2x+5(sinx+cosx)=0

Решите задачу:

$sin2x+5(sinx+cosx)=0\\\\t=sinx+cosx\; ,\\\\ t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x,\\\\sin2x=t^2-1\\\\t^2+5t-1=0\\\\D=25+4=29\; ,\; t_{1,2}= \frac{-5\pm \sqrt{29}}{2} \\\\t_1\approx-5,19\; \; ;\; \; t_2\approx0,19\\\\a)\quad sinx+cosx= \frac{-5-\sqrt{29}}{2} \, \left |:\sqrt2$

$\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}cosx= \frac{-5-\sqrt{29}}{2\sqrt2} \\\\cos\frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx=\frac{-5-\sqrt{29}}{2\sqrt2}\\\\\frac{-5-\sqrt{29}}{2\sqrt2}\approx -3,71$

$sin(\frac{\pi}{4}+x)\approx -3,71\ \textless \ -1\; ,\; \; a\; \; -1 \leq sin(\frac{\pi}{4}+x) \leq 1\\\\net\; \; reshenij\\\\b)\quad sinx+cosx=\frac{-5+\sqrt{29}}{2}|:\sqrt2 \\\\sin(\frac{\pi}{4}+x)= \frac{-5+\sqrt{29}}{2\sqrt2}}\\\\\frac{\pi}{4}+x=(-1)^{n}\cdot arcsin\frac{-5+\sqrt{29}}{2\sqrt2}+\pi n,\; \; n\in Z$

$x=-\frac{\pi}{4}+(-1)^{n}\cdot arcsin\frac{-5+\sqrt{29}}{2\sqrt2}+\pi n$