Pro tip: It's better to turn these into the form cos x = cos a which has solutions x = \pm a + 360^circ k quad for integer k.
This one is already about 2x so it's easier to leave it like that.
Linear combinations of sine and cosine of the same angle are phase shifted cosines.
3 sin(2x) + 2 cos(2x) = 3
\sqrt{13} ( 2/ sqrt{13} cos(2x) + 3/sqrt{13) sin(2x) ) = 3
2/ sqrt{13} cos(2x) + 3/sqrt{13) sin(2x) = 3/sqrt{13}
Let's let theta = arctan(3/2) approx 56.31 ^circ
We really mean the one in the first quadrant.
(If we wanted to do sine instead of cosine like we're doing, we would use arctan(2/3) .)
We have cos theta = 2/sqrt{13} and sin theta = 3/sqrt{13}.
cos theta cos(2x) + sin theta sin(2x) = sin theta
cos(2x - theta) = cos (90^circ - theta)
2x - theta = \pm (90^circ - theta) + 360^circ k
2x = theta \pm (90^circ - theta) + 360^circ k
x = theta/2 \pm (45^circ - theta/2) + 180^circ k
x = 45^circ + 180^circ k or x = theta - 45^circ + 180^circ k
x = 45^circ + 180^circ k or x = arctan(3/2) - 45^circ + 180^circ k
Since 56.31-45 = 11.31
x = 45^circ + 180^circ k or x approx 11.31^circ + 180^circ k
