Question

3sin^2 (2x)+sin2x=(sinx-cosx)^2

Answer 1

3sin²2x+sin2x=sin²x-2sinxcosx+cos²x
3sin²2x+sin2x=1-sin2x
3sin²2x+sin2x-1+sin2x=0
3sin²2x+2sin2x-1=0
sin2x=a
3a²+2a-1=0
D=4+12=16
a1=(-2-4)/6=-1⇒sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn
a2=(-2+4)/6=1/3⇒sin2x=1/3⇒2x=(-1)^n*arcsin1/3+πn⇒x=(-1)^n*1/2arcsin1/3+πn/2