Question

6cos^2x=9cosx-4sin^2x

Answer 1

$\mathtt{6cos^2x=9cosx-4sin^2x;~6cos^2x=9cosx-4(1-cos^2x);~}\\\mathtt{2cos^2x-9cosx+4=0;~2(cosx-\frac{1}{2})(cosx-\frac{8}{2})=0;~cosx=\frac{1}{2}}$

ОТВЕТ: $\mathtt{x=\frac{\pi}{3}+2\pi k,~\frac{5\pi}{3}+2\pi n;~k,n\in Z}$