Sin2x-sin(3п/2-2x)+1=0

Question

Sin2x-sin(3п/2-2x)+1=0

Comments

sin2x+cos2x+1=02sinxcosx+cos²x-sin²x+sin²x+cos²x=02sinxcosx+2cos²x=02сosx(sinx+cosx)=01) cosx=0x=пn2)sinx+cosx=0/cosxtgx=-1x=-п/4+пk

Answer 1

Sin2x+cos2x+1=0
2sinxcosx+cos²x-sin²x+sin²x+cos²x=0
2sinxcosx+2cos²x=0
2cosx(sinx+sin(π/2-x)=0
2cosx*2sinπ/4cos(x-π/4)=0
2*2*√2/2cosxcos(x-π/4)=0
2√2cosxcos(x-π/4)=0
cosx=0⇒x=π/2+πn
cos(x-π/4)=0⇒x-π/4=π/2+πn⇒x=3π/4+πn

Comments

sin2x+cos2x+1=0
2sinxcosx+cos²x-sin²x+sin²x+cos²x=0