Solve the equation sin(3x) + sin^2 (x)= 2

help!!!
please show the step~~

sin(3x) + sin^2 (x) = 2
sin(2x)cos(x) + sin(x)cos(2x) + sin^2(x) = 2
2sin(x)cos^2(x) + sin(x)cos^2(x) - sin^3(x) + sin^2(x) = 2
3sin(x)(1 - sin^2(x)) - sin^3(x) + sin^2(x) = 2
3sin(x) - 3sin^3(x) - sin^3(x) + sin^2(x) = 2
let y = sin(x) for simplicity
4y^3 - y^2 - 3y + 2 = 0
y = -1 is seen to be a solution so

(y + 1)(4y^2 - 5y + 2) = 0

we see that discriminant of 4y^2 - 5y + 2 is less than zero
so the other two solutions are complex. we ignore for our eqn.

so we now solve sin(x) = -1 in the range
ie x = 270 degrees

OK.
Sin(3x) = Sin(2x+x) = cos(2x)*sin(x) + sin(2x)*cos(x)
cos(2x) = 1-2*sin^2(x) and sin(2x) = 2sin(x)*cos(x)
So, you have

sin(x)-2sin^3(x) +2sin(x)*cos^2(x) + sin^2(x) = 2 Now cos^2(x) = 1-sin^2(x)
so you have sin(x) -2sin^3(x) +2sin(x) -2sin^3(x) +sin^2(x) = 2.
so now you have -4sin^3(x) +sin^2(x) +3sin(x) -2 = 0

Trying to solve this analytically is quite cumbersome.
I looked at the graph of this and there appears to be a solution in the interval you provided at
3pi/2 radians or 270 degrees.

sin 3x = sin ( x + 2x ) = sin x cos 2x + cos x sin 2x