Question

Пожалуйста, помогите решить уравнение

Answer 1

$cos^{2} x-2sin2x+3sin^{2} x=0\\2sin^{2} x-4sinxcosx+sin^{2} x+cos^{2} x=0\\3sin^{2} x-4sinxcosx+cos^{2} x=0$

$3tg^{2} x-4tgx+1=0\\D=b^{2} -4ac=(-4)^{2} -4*3*1=16-12=4=2^{2} \\tg_{1} =\frac{-b+\sqrt{D} }{2a} = \frac{4+2}{2*3} = \frac{6}{6} = 1\\tg_{2} =\frac{-b-\sqrt{D} }{2a} = \frac{4-2}{2*3} = \frac{2}{6} = \frac{1}{3}$

$tgx=1\\tgx=\frac{1}{3}$

При tgx=1

$x=\frac{\pi }{4} +\pi n,n$∈Z

При tgx=$\frac{1}{3}$

$x=arctg\frac{1}{3} +\pi n,n$∈Z

Ответ:

$x=\frac{\pi }{4} +\pi n,n$∈Z

$x=arctg\frac{1}{3} +\pi n,n$∈Z