Question

How to solve this?`sinx+sin2x+sin3x=0`

Answer 1

`x = (kpi)/2`
`x = +- (2pi)/3 + 2kpi`

Explanation:

Use trig identity:
`sin a + sin b = 2sin ((a + b)/2).cos ((a - b)/2)`
In this case:
`sin x + sin 3x = 2sin (2x).cos (x)`
`(sin x + sin 3x) + sin 2x = 2sin (2x)cos (x) + sin (2x) =`
`= sin (2x)(2cos x + 1) = 0`
Either one of the 2 factors must be zero.
a. sin 2x = 0
2x = 0 --> x = 0
`2x = pi` --> `x = pi/2`
`2x = 2pi` --> `x = pi`
General answer: `x = (kpi)/2`
b. 2cos x + 1 = 0 --> `cos x = - 1/2`
Trig table and unit circle give 2 solutions -->
`x = +- (2pi)/3 + 2kpi`