How do you solve #cos2x+2sinx=0# using the double angle identity?

1 Answer
Aug 14, 2015

Solve f(x) = cos 2x + 2sin x = 0

Ans: 201.47 and 338.53 deg

Explanation:

Replace #cos 2x# by #1 - 2sin^2 x#
#1 - 2sin^2 x + 2sin x = 0#
#2sin ^2 x - 2sin x - 1 = 0#
#D = d^2 = b^2 - 4ac = 4 + 8 = 12# --> #d = +- 2sqrt3#
#sin x = 1/2 +- sqrt3/2#
Only the negative answer is accepted
#sin x = (1 - sqrt3)/2 = - 0.732/2 = 0.366# --> #x = - 21.47# deg.
Trig unit circle gives 2 answers within interval (0, 2pi):
#x = 180 + 21.47 = 201.47# and #x = 360 - 21,47 = 338.53# deg

Check: x = 201,47 --> 2x = 402.94 = 42.94 + 360 --> cos 2x = cos 42.94 = 0.73 --> 2sin 201.47 = -0.73.
cos 2x - 2sin x = 0.73 - 0.73 = 0. OK