Question

cos^2x-3sin^2x=-sin2x

Answer 1

Cos^2x-3sin^2x=-sin2x   
1-sin²2x-3sin²2x+sin2x=0
-4sin²2x+sin2x+1=0
4sin²2x-sin2x -1=0  √D=√1+16=√17
sin2x=1/8[1+√17]     x=1/2(-1)ⁿarcsin[1/8[1+√17] ]+πn/2
sin2x=1/8[1-√17]     x=1/2(-1)ⁿarcsin[1/8[1-√17] ]+πn/2