Question

Решите , 1)2sin^2 x -3sin2x-4cos2x=4 ^2-это в квадрат 2)4sin2x=4-cos2x

Answer 1

$2sin^2x-3sin2x-4cos2x=4\\2sin^2x-6sinxcosx-4cos^2x+4sin^2x=4(sin^2x+cos^2x)\\6sin^2x-6sinxcosx-4cos^2x-4sin^2x-4cos^2x=0\\2sin^2x-6sinxcosx-8cos^2x=0|\; :2cos^2x\\tg^2x-3tgx-4=0\\tgx=u\\u^2-3u-4=0\\D:\; 9+16=25\\u=\frac{3\pm 5}{2}\\\\u_1=4\\tgx=4\\x=arctg4 + \pi n;\\\\u_2=-1\\tgx=-\frac{\pi}{4}+\pi n,\; n\in Z$

$4sin2x=4-cos2x\\8sinxcosx=4(sin^2x+cos^2x)-(cos^2x-sin^2x)\\8sinxcosx-4sin^2x-4cos^2x+cos^2x-sin^2x=0\\-5sin^2x-3cos^2x+8sinxcosx=0|\; :(-cos^2x)\\5tg^2x-8tgx+3=0\\tgx=u\\5u^2-8u+3=0\\D:\; 64-60=4\\u=\frac{8\pm 2}{10}\\\\u_1=1\\tgx=1\\x=\frac{\pi}{4}+\pi n,\; n \in Z; \\\\u_2=\frac{3}{5}\\tgx=\frac{3}{5}\\x=arctg \frac{3}{5}+\pi n, \;n\in Z$