Question

Помогите решить пожалуйста
${ \cos^{2}x } + \cos^{2}2x - \cos {}^{2} 3x - \cos {}^{2} 4x = 0$
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Answer 1

$Cos^{2}x +Cos^{2}2x-Cos^{2}3x-Cos^{2}4x =0\\\\\\\dfrac{1+Cos2\alpha }{2} +\dfrac{1+Cos4\alpha }{2}-\dfrac{1+Cos6\alpha }{2}-\dfrac{1+Cos8\alpha }{2}=0\\\\\\1+Cos2\alpha+1+Cos4\alpha -1-Cos6\alpha -1-Cos8\alpha =0 \\\\\\Cos2\alpha+Cos4\alpha -Cos6\alpha -Cos8\alpha =0\\\\\\(Cos2\alpha-Cos8\alpha)+(Cos4\alpha-Cos6\alpha)=0$

$-2Sin\dfrac{2\alpha-8\alpha}{2}Sin\dfrac{2\alpha+8\alpha}{2}-2Sin\dfrac{4\alpha-6\alpha}{2}Sin\dfrac{4\alpha+6\alpha}{2} =0\\\\Sin3\alpha Sin5\alpha+Sin\alpha Sin5\alpha=0\\\\Sin5\alpha (Sin3\alpha+Sin\alpha)=0\\\\\\Sin5\alpha\cdot2Sin\dfrac{3\alpha +\alpha }{2} Cos\dfrac{3\alpha-\alpha}{2} =0\\\\\\Sin5\alpha Sin2\alpha Cos\alpha=0$

$\left[\begin{array}{ccc}Sin5\alpha=0 \\Sin2\alpha=0 \\Cos\alpha=0 \end{array}\right\\\\\\\left[\begin{array}{ccc}5\alpha=\pi n,n\in Z \\2\alpha=\pi n,n\in Z \\\alpha=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}\alpha _{1}=\dfrac{\pi n }{5} ,n\in Z \\\alpha _{2}=\dfrac{\pi n }{2},n\in Z \\\alpha _{3}=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right \\\\\\Otvet:\boxed{\dfrac{\pi n }{5} \ ; \ \dfrac{\pi n }{2},n\in Z}$