Question

Решите пожалуйстаа!!
sin10x-2cos5x=0 и
cos6x-cos4x=2sinx

Answer 1

$sin10x-2cos5x=0\ 2sin5xcos5x-2cos5x=0 \ 2cos5x(sin5x-1)=0 \ left[begin{array}{ccc}cos5x=0\sin5x=1end{array}rightto left[begin{array}{ccc}x_1= frac{ pi }{10}+ frac{pi n}{5}, n in Z \ x_2= frac{pi}{10}+ frac{2pi k}{5} , k in Z end{array}right$

$cos 6x-cos 4x =2sin x \ 2sin 5xsin x-2sin x=0 \ 2sin x(sin 5x-1)=0 \ left[begin{array}{ccc}sin x=0\sin 5x=1end{array}rightto left[begin{array}{ccc}x_1=pi k,k in Z\ x_2= frac{pi}{10} + frac{2pi k}{5}, k in Z end{array}right$

Answer 2

Спасибо огромное :)