Question

3sin^2x-4sinxcosx+5sin^2((п/2)+x)=2

Answer 1

3sin^2х-4sinxcosx+5cos^2x-2=0

3sin^2х-4sinxcosx+5cos^2x-2•(sin^2x+cos^2x)=0

3sin^2х-4sinxcosx+5cos^2x-2sin^2x-2cos^2x=0

Sin^2x-4sinxcosx+3cos^x=0. (:cos^2x)

tg^2x-4tgx+3=0

tgx=t

t^2-4t+3=0

D=16-12=4

t1=(4+2):2=3

t2=(4-2):2=1

tgx=1

X=arctg1+Пк;к€Z

X=П/4+Пк;к€Z

tgx=3

X=arctg3+Пк;к€Z