Question

Sin^2x- корень из 2*cos(2x-pi/4)=1

Answer 1

Sin²x-√2cos(2x-π/4)=1
Sin²x-√2cos(2x-π/4)-1=0
-cos²x-√2cos(2x-π/4)=0
(1+cos2x)/2+√2cos(2x-π/4)=0
1+cos2x+2√2cos2xsinπ/4+2√2sin2xcosπ/4=0
1+cos2x+2√2cos2x*√2/2+2√2sin2x*√2/2=0
1+cos2x+2cos2x+2sin2x=0
Sin²x+cos²x+3cos²x-3sin²x+4sinxcosx=0
2sin²x-4sinxcosx-4cos²x=0 /2cos²x≠0
tg²x-2tgx-2=0
tgx=a
a²-2a-2=0
D=4+8=12
a1=(2-2√3)/2=1-√3⇒tgx=1-√3⇒x=arctg(1-√3)+πn
a2=(2+2√3)/2=1+√3⇒tgx=1+√3⇒x=arctg(1+√3)+πn