Trigonometry Help Needed, 10 points for best solution

Solve the following equation for 0°≤x≤360°

sin 3x + 3 cos 2x + sin x = 3

please show a complete working for me to understand,
Thank you!

sin 3x + 3 cos 2x + sin x = 3
=> 3sinx - 4sin^3 x + 3 (1 - 2sin^2 x) + sinx = 3
=> 4sin^3 x + 6sin^2 x - 4sinx = 0
=> sinx (2sin^2 x + 3sinx - 2) = 0
=> sinx (2sinx - 1)(sinx + 2) = 0
=> sinx = 0 or 1/2 or - 2 ... (sinx = - 2 is not possible)
=> x = {0, 30°, 150°, 180°, 360°}.

we know that sin3x = 3sinx - 4sin^3x and cos2x = 1-2sin^2x
=> sin 3x + 3 cos 2x + sin x = 3sinx - 4sin^3x + 3-6sin^2x + sinx
=> 4sinx - 4sin^3x + 3 - 6sin^2x = 3
=> 4sinx - 4sin^3x - 6sin^2x = 0
=> 2sinx - 2sin^3x - 3sin^2x = 0
=> sinx(2-2sin^2x - 3sinx) = 0
=> sinx = 0 or 2-2sin^2x - 3sinx = 0
if sinx =0 then x = 0, pi, or 2pi
Now if 2-2sin^2x - 3sinx = 0
=> 2sin^2x + 3sinx - 2 = 0
=> 2sin^2x + 4sinx - sinx -2 = 0
=> 2sinx(sinx + 2) - 1(sinx+2) = 0
=> (sinx + 2) (2sinx -1) = 0
=> sinx + 2 = 0 or 2sinx -1 = 0
=> sinx = -2 or sinx = 1/2
sinx = -2 is not possible because value of sin remains between (-1,1)
so sinx = 1/2 => x = arcsin(1/2)
=> x = pi/6 or 5pi/6
=> x = (0,pi, 2pi, pi/6, 5pi/6)
=>

My friend gave the good answer. I have nothing more to explain. But give me that 10 points please!