Question

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

Answer 1

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

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

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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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