Question

Решить примеры: 1) sin7x=sinx 2) √2sinx/2≥1 3) (sinx+cosx)^2=cos2x 4) 3sin2x+4cos2x=5

Answer 1

$2.\quad \sqrt2sin\frac{x}{2} \geq 1\\\\sin\frac{x}{2} \geq \frac{1}{\sqrt2}\\\\\frac{\pi}{4}+2\pi n \leq \frac{x}{2} \leq \frac{3\pi }{4}+2\pi n\\\\\frac{\pi}{2}+4\pi n \leq x \leq \frac{3\pi }{2}+4\pi n,\quad n\in Z$

$4.\quad 3sin2x+4cos2x=5\; |:5\; \; (5=\sqrt{3^2+4^2}=\sqrt{25})\\\\\frac{3}{5}sin2x+\frac{4}{5}cos2x=1\\\\(\; \; (\frac{3}{5})^2+(\frac{4}{5})^2=1\; \; \to \frac{3}{5}=sin \alpha ,\; \frac{4}{5}=cos \alpha \to tg \alpha =\frac{3}{4},\alpha =arctg\frac{3}{4})\\\\sin \alpha \cdot sin2x+cos \alpha \cdot cos2x=1\\\\sin( 2x+ \alpha )=1\\\\2x+ \alpha =\frac{\pi}{2}+2\pi n\\\\2x=\frac{\pi}{2}- \alpha +2\pi n\\\\x=\frac{\pi}{4}-\frac{ \alpha }{2}+\pi n,\; n\in Z\\\\x=\frac{\pi}{4}-\frac{arctg\frac{3}{4}}{2}+\pi n$

Answer 2

Решение  1,3,4  во вложении.