How do you solve #sin2x+sinx+2cosx+1=0# in the interval [0,360]?

1 Answer
Aug 22, 2016

#120^@,240^@, 270^@#

Explanation:

Use trig identity:
sin 2x = 2sin x.cos x.
Replace in the equation sin 2x by 2sin x.cos x:
2sin x.cos x + sin x + 2cos x + 1 = 0
Proceed factoring by grouping:
sin x(2cos x + 1) + 2cos x + 1 = 0
(2cos x + 1)(sin x + 1) = 0
a. 2cos x + 1 = 0 --> #cos x = -1/2#
Trig table and unit circle -->
#cos x = -1/2# --> #x = +- 120^@#
Arc #- 120^@# is co-terminal to arc #240^@#
b. sin x + 1 = 0 --> sin x = -1
sin x = - 1 --> #x = 270^@#
Answers for (0, 360):
#120^@; 240^@, 270^@#