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#
