Question

sin^2x-2cos^2x=sin2x

Answer 1

$sin^2x-2cos^2x=sin2x \ sin^2x-2cos^2x-2sinxcosx=0$
делим все уравнение на cos²x
$tg^2x-2tgx-2=0 \ frac{D}{4}=1+2=3 \ tgx_1=1+ sqrt{3} \ tgx_2=1- sqrt{3} \ \left[begin{array}{I} x=arctg(1+ sqrt{3})+ pi k \ x=arctg(1- sqrt{3})+ pi k end{array}} ; k in Z$

Ответ: $left[begin{array}{I} x=arctg(1+ sqrt{3})+ pi k \ x=arctg(1- sqrt{3})+ pi k end{array}} ; k in Z$