Question

1. Упростить
2. Тригонометрическое уравнение

Answer 1

$1)\sqrt{32}Sin\frac{33\pi }{4}-Cos^{2}\frac{23\pi }{16}=\sqrt{32}Sin(8\pi+\frac{\pi }{4})-Cos^{2}(\frac{3\pi }{2}-\frac{\pi }{16})=\sqrt{32}Sin\frac{\pi }{4}-Sin^{2}\frac{\pi }{16}=\sqrt{32}*\frac{\sqrt{2}}{2}-Sin^{2}\frac{\pi }{16}=\frac{\sqrt{64} }{2}-Sin^{2}\frac{\pi }{16}=4-Sin^{2}\frac{\pi }{16}$

$2)(2Sin^{2}x-5Sinx-3)*\sqrt{-2Cosx}=0$

ОДЗ :

-2Cosx ≥ 0

Cosx ≤ 0

$x\in[\frac{\pi }{2}+2\pi n;\frac{3\pi }{2}+2\pi n],n\in Z\\\\1)2Sin^{2}x-5Sinx-3=0\\\\Sinx=m,-1\leq m\leq1\\\\2m^{2}-5m-3=0\\\\D=(-5)^{2}-4*2*(-3)=25+24=49=7^{2}\\\\m_{1}=\frac{5+7}{4}=3-neyd\\\\m_{2}=\frac{5-7}{4}=-\frac{1}{2}\\\\Sinx=-\frac{1}{2}\\\\x=-\frac{\pi }{6}+2\pi n,n\in Z-neyd...odz\\\\x=\frac{7\pi }{6}+2\pi n,n\in Z \\\\2)-2Cosx=0\\\\Cosx=0\\\\x=\frac{\pi }{2}+\pi n,n\in Z \\\\Otvet:\boxed{\frac{\pi }{2}+\pi n;\frac{7\pi }{6}+2\pi n,n\in Z}$