A quite hard Question

For 0=< x =< 360
Solve 3 sin x + 2 cos x = 2
3 sin x – 2 = - 2 cos x
9 sin2x – 12sinx + 4 = 4cos2x
9 sin2x – 12sinx + 4 = 4 – 4sin2x
13 sin2x – 12sinx + 0 = 0
sin x(13sinx – 12) = 0
sin x = 0 或 13sin x – 12 = 0
sinx = 0
x = 0 或 180o
13sin x – 12 = 0
當 sin x = 12/13
x = 67.4o
或
x = 112.6o

let theta=arctan(2/3)=33.7deg (choose 1st quadrant)
then cos(theta)=3/sqrt(2^2+3^2)=3/sqrt(13) and sin(theta)=2/sqrt(13)
(sqrt = square root of)

divide the equation by sqrt(13)
3/sqrt(13) sinx + 2/sqrt(13) cosx = 2/sqrt(13)
cos(theta)sin(x) + sin(theta)cos(x)=sin(theta)
sin(x+theta)=sin(theta)

x+theta=theta
=&gt; x = 0deg, 360deg

or x+theta=180deg-theta
=&gt; x=180deg-2*theta=180-2*33.7 = 112.6deg

x = 67.4deg is wrong
3sin(67.4deg)+2cos(67.4deg) = 3.538

x = 180deg is wrong
3sin(180deg)+2cos(180deg) = -2