Question

3cos²x=4sinx·cosx-sin²x

Answer 1

$4cos^2x=4sinxcosx-sin^2x(:cos^2x) \ 4=4tgx-tg^2x \ tg^2x-4tgx+4=0 \ (tgx-2)^2=0 \ tgx-2=0 \ tgx=2 \ x=arctg2+ pi k$

x=arctg2+pi*k, k∈Z