Question

Решите пожалуйста
2sin(3pi/2 +x)cos(pi/2+x)=√3cos(2pi-x)
найти корни на промежутке [-2pi;-pi]
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Answer 1

$2Sin(\frac{3\pi }{2}+x)Cos(\frac{\pi }{2}+x)=\sqrt{3}Cos(2\pi-x)\\\\-2Cosx*(-Sinx)=\sqrt{3}Cosx\\\\2Sinx Cosx-\sqrt{3}Cosx=0\\\\Cosx(2Sinx-\sqrt{3})=0\\\\\left[\begin{array}{ccc}Cosx=0\\Sinx=\frac{\sqrt{3}}{2} \end{array}\right$

$1)Cosx=0\\\\x=\frac{\pi }{2}+\pi n,n\in Z\\\\n=-2\Rightarrow x=\frac{\pi }{2}-2\pi= -\frac{3\pi }{2}$

$2)Sinx=\frac{\sqrt{3}}{2}\\\\x_{1}=\frac{\pi }{3}+2\pi n,n\in Z \\\\n=-1\Rightarrow x=\frac{\pi }{3}-2\pi=-\frac{5\pi }{3}\\\\x_{2}=\frac{2\pi }{3}+2\pi n,n\in Z\\\\n=-1\Rightarrow x=\frac{2\pi }{3}-2\pi=-\frac{4\pi }{3}\\\\Otvet:\boxed{-\frac{3\pi }{2};-\frac{5\pi }{3};-\frac{4\pi }{3}}$