Question

помогите решить (1+tgx)(1-sin2x)=1-tgx

Answer 1

$(1+tgx)(1-sin2x)=1-tgx$

ООУ:$x neq frac{pi}{2} + pi t,t in Z$

$(1+tgx)(1-2sinxcosx)=1-tgx$

$1+tgx-2sin^2x-2sinxcosx=1-tgx$

$tgx-sin^2x-sinxcosx=0$

$sinx(frac{1}{cosx}-sinx-cosx)=0$

$tgx(1-sinxcosx-cos^2x)=0$

$tgx(sin^2x+cos^2x-sinxcosx-cos^2x)=0$

$tgxsinx(sinx-cosx)=0$

$x=pi k,k in Z$

$x=frac{pi}{4}+pi n,n in Z$