Question

Решите пожалуйста 1 и 3

Answer 1

$\mathtt{\frac{sin\alpha}{1+cos\alpha}-\frac{sin\alpha}{1-cos\alpha}=\frac{1-cos\alpha-1-cos\alpha}{sin\alpha}=\frac{-2cos\alpha}{sin\alpha}=-2ctg\alpha}$

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

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