Question

Решите уравнение
Sin2x+2cos^2x+cos2x=0

Answer 1

$\sin(2x) + 2\cos^2x+\cos(2x)=0\\2\sin x\cos x+2\cos^2 x+\cos^2x-\sin^2x=0\\(\cos^2x+2\sin x\cos x+\sin^2x)+2(\cos^2x-\sin^2x)=0\\(\cos x+\sin x)^2+2(\cos x+\sin x)(\cos x-\sin x)=0\\(\cos x+\sin x)(\cos x+\sin x+2\cos x-2\sin x)=0\\(\cos x+\sin x)(3\cos x-\sin x)=0$

$1)~\cos x+\sin x=0~~~|:\cos x\neq 0\\\\~~~\dfrac {\cos x}{\cos x}+\dfrac {\sin x}{\cos x}=0;~~1+tg~x=0;~~tg~x=-1\\\\~~~~\boxed{\boldsymbol{x_1=-\dfrac{\pi}4+\pi n,~n\in \mathbb Z}}\\\\2)~3\cos x-\sin x=0~~~|:\cos x\neq 0\\\\~~~\dfrac {3\cos x}{\cos x}-\dfrac {\sin x}{\cos x}=0;~~~3-tg~x=0; ~~~tg~x=3\\\\~~~~\boxed{\boldsymbol{x_2=arctg~3+\pi k,~k\in \mathbb Z}}$

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Использованы формулы :

sin (2α) = 2 sin α cos α

cos (2α) = cos²α - sin²α

a² + 2ab + b² = (a + b)²

a² - b² = (a + b) (a - b)

Answer 2

Решение приложено

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