Question

20б! решите: sin²x+sin²2x+sin²3x=1,5

Answer 1

Пользуемся формулой понижения степени:
$cos2x=cos^2x-sin^2x=(1-cos^2x)-sin^2x=1-2sin^2x\\cos2x=1-2sin^2x\\\frac{1-cos2x}{2}=sin^2x;\\\\1,5=1\frac{1}{2}=\frac{3}{2};$

$sin^2x+sin^22x+sin^23x=1,5\\\frac{1-cos2x}{2}+\frac{1-cos4x}{2}+\frac{1-cos6x}{2}=\frac{3}{2}\\\frac{1-cos2x+1-cos4x+1-cos6x-3}{2}=0|*2\\-cos2x-cos4x-cos6x=0\\-(cos2x+cos6x)-cos4x=0\\-(2cos4xcos2x)-cos4x=0\\-cos4x(2cos2x-1)=0\\\\cos4x=0\\4x=\frac{\pi}{2}+\pi n\\x=\frac{\pi}{8}+\frac{\pi n}{4}, \; n\in Z;\\\\2cos2x-1=0\\cos2x=\frac{1}{2}\\2x=\frac{\pi}{3}+2\pi n\\x=\frac{\pi}{6}+\pi n, \; n\in Z.$