Question

4sin^2 3x-sin3x=2+sin^2 3x

Answer 1

4sin²3x-sin3x=2+sin²3x
3sin²3x-sin3x-2=0
z=sin3x
3z²-z-2=0   D=1+24  √D=√25=5   z1=1/6[1+5]=1   z2=1/6[1-5]=-4/6=-2/3

sin3x=1  3x=π/2+2πn   x=π/6+2/3*πn   n∈Z
sin3x=-2/3   3x=(-1)ⁿarcsin(-2/3)+πn    x=(-1)ⁿ*1/3*arcsin(-2/3)+πn/3π  n∈Z