Помогите решить 140,141

Question 1

Question 2

Решите задачу:

$140.\quad 1+2cos^2x+2\sqrt2\cdot sinx+cos2x=0\\\\\star \; \; cos^2x=1-sin^2x\; \; \star \\\\\star \; \; cos2x=cos^2x-sin^2x=(1-sin^2x)-sin^2x=1-2sin^2x\; \; \star \\\\\\1+2-2sin^2x+2\sqrt2\cdot sinx+1-2sin^2x=0\\\\-4sin^2x+2\sqrt2\cdot sinx+4=0\; |:(-2)\\\\2sin^2x-\sqrt2\cdot sinx-2=0\\\\D=2+4\cdot 2\cdot 2=18\; ,\\\\a)\; \; (sinx)_1= \frac{\sqrt2-\sqrt{18}}{4}= \frac{\sqrt2-3\sqrt2}{4}=\frac{-2\sqrt2}{4}=-\frac{\sqrt2}{2}\approx -0,71$

$x=(-1)^{n}arcsin(-\frac{\sqrt2}{2})+\pi n=(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n,\; n\in Z$

$b)\; \; (sinx)_2= \frac{\sqrt2+\sqrt{18}}{4} = \frac{4\sqrt2}{4}=\sqrt2\approx 1,41\ \textgreater \ 1\\\\net\; \; reshenij\\\\Otvet:\; \; x=(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n,\; n\in Z$

$141.\quad cos(2x+ \frac{2\pi }{3})+4sin(x+\frac{\pi }{3})=2,5\\\\\star \; \; 2x+ \frac{2\pi }{3} =2\cdot (x+\frac{\pi}{3})\; \to \; \alpha =x+\frac{\pi}{3}\; ,\; \; 2 \alpha =2x+\frac{2\pi}{3}\; \star \\\\\star \; \; cos2 \alpha =1-2sin^2 \alpha \; \; \star \\\\\\cos2 \alpha +4sin \alpha =2,5\\\\1-2sin^2 \alpha +4sin \alpha -2,5=0\; |\cdot (-1)\\\\2sin^2 \alpha -4sin \alpha +1,5=0\; |\cdot 2\\\\4sin^2 \alpha -8sin \alpha +3=0\\\\D/4=4^2-4\cdot 3=4\; ,\\\\a)\; \; (sin \alpha )_1= \frac{4-2}{4}=\frac{1}{2}$

$\alpha =(-1)^{n}\cdot \frac{\pi}{6}+\pi n,\; n\in Z\\\\x+\frac{\pi}{3}=(-1)^{n}\cdot \frac{\pi}{6}+\pi n,\; n\in Z\\\\x=-\frac{\pi}{3}+(-1)^{n}\cdot \frac{\pi}{6}+\pi n,\; n\in Z\\\\b)\; \; \; (sin \alpha )_2=\frac{4+2}{4}= \frac{3}{2}\ \textgreater \ 1\; \; \to \; \; net\; reshenij \\\\Otvet :\; \; x=-\frac{\pi}{3}+(-1)^{n}\cdot \frac{\pi}{6}+\pi n,\; n\in Z$

$P.S.\; \; ax^2+2px+q=0\; ,\quad (2p\, \vdots \, 2)\\\\D/4=p^2-aq\ \; ,\; \; x_{1,2}= \frac{-p\pm \sqrt{D/4}}{a}$