Question

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

Answer 1

`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^@`