Question

Для егэ решить, утром профильная у сестры ! 1) корень из 6cos x+2sin (2x+п/3)+корень из 3=sin2x [3п; 9п/] 2) cos x + корень из 2 sin (2x+п/4)+1=sin 2x [-11п/2; -4п]

Answer 1

$\sqrt{6}Cosx+2Sin(2x+ \frac{ \pi }{3})+ \sqrt{3}=Sin2x\\\\ \sqrt{6} Cosx+2(Sin2xCos \frac{ \pi }{3}+Sin \frac{ \pi }{3}Cos2x)+ \sqrt{3} =Sin2x\\\\ \sqrt{6}Cosx+2( \frac{1}{2} Sin2x+ \frac{ \sqrt{3} }{2}Cos2x)+ \sqrt{3}-Sin2x=0\\\\ \sqrt{6} Cosx+Sin2x+ \sqrt{3}Cos2x+ \sqrt{3}-Sin2x=0\\\\ \sqrt{6}Cosx+ \sqrt{3}Cos2x+ \sqrt{3} =0\\\\ \sqrt{2} Cosx+Cos2x+1=0\\\\ \sqrt{2}Cosx+2Cos ^{2}x-1+1=0\\\\ \sqrt{2} Cosx+2Cos ^{2}x=0\\\\Cosx+ \sqrt{2} Cos ^{2} x =0 \\\\ Cosx(1+ \sqrt{2} Cosx)=0 \\\\Cosx=0$
$x= \frac{ \pi }{2}+ \pi n\\\\\\ \sqrt{2}Cosx=-1\\\\Cosx=- \frac{1}{ \sqrt{2} }\\\\x=+- \frac{3 \pi }{4}+2 \pi n\\\\\\3 \pi \leq \frac{ \pi }{2} + \pi n \leq \frac{9 \pi }{2} \\\\3 \leq \frac{1}{2} +n \leq \frac{9}{2} \\\\ \frac{5}{2} \leq n \leq \frac{8}{2}\\\\n=3\\\\ x_{1} =\frac{ \pi }{2} +3 \pi= \frac{7 \pi }{2}\\\\n=4\\\\ x_{2}= \frac{ \pi }{2} +4 \pi = \frac{9 \pi }{2} \\\\\\3 \pi \leq \frac{3 \pi }{4}+2 \pi n \leq \frac{9 \pi }{2} \\\\3 \leq \frac{3}{4}+2n \leq \frac{9}{2}$

$\frac{9}{4} \leq 2n \leq \frac{15}{4}\\\\ \frac{9}{8} \leq n \leq \frac{15}{8}\\\\\\3 \pi \leq - \frac{3 \pi }{4}+2 \pi n \leq \frac{9 \pi }{2} \\\\3 \leq - \frac{3}{4} +2n \leq \frac{9}{2}\\\\ \frac{15}{4} \leq 2n \leq \frac{21}{4}\\\\ \frac{15}{8} \leq n \leq \frac{21}{8}\\\\\\n=2 \\\\ x_{3} =- \frac{3 \pi }{4}+4 \pi = \frac{13 \pi }{4}$