Question

Решите пожалуйста 1 и 3

Answer 1

$mathtt{frac{sinalpha}{1+cosalpha}-frac{sinalpha}{1-cosalpha}=frac{1-cosalpha-1-cosalpha}{sinalpha}=frac{-2cosalpha}{sinalpha}=-2ctgalpha}$

$mathtt{cos(frac{pi}{6}-x)-frac{sqrt{3}}{2}cosx=frac{1}{4};~cosfrac{pi}{6}cosx+sinfrac{pi}{6}sinx-frac{sqrt{3}}{2}cosx=frac{1}{4};}\\mathtt{frac{1}{2}sinx=frac{1}{4};~sinx=frac{1}{2};~x=(-1)^kfrac{pi}{6}+pi k,kin Z}$

$mathtt{3(1-cos^2x)+7cosx-5=0;~3cos^2x-7cosx+2=0;~}\\mathtt{left[begin{array}{ccc}mathtt{cosx=2}\mathtt{cosx=frac{1}{3}}end{array}right;~cosx=frac{1}{3};~x=бarccosfrac{1}{3}+2pi k,kin Z}$