Question

How you solve this? to solve and discuss after real parameter "a" the equation:`sinx+asin2x+sin3x=0`

Answer 1

See below.

Explanation:

Using and substituting

`sin(3x)=3cos^2xsinx-sin^3x` and
`sin(2x)=2cosx sinx`

`3cos^2xsinx-sin^3x+sinx+2asinxcosx=0`

or

`(3cos^2x-sin^2+1+2acosx)sinx=0`

or

`(4cos^2x+2acosx)sinx=0`

or

`(4cosx+2a)cosx sinx = 0`

so we have the possibilities

`{(cosx=0),(sinx=0),(cosx+a/2=0):}`

with the outcomes

`(x=pm pi/2+2kpi) uu (x=pm pi + 2kpi) uu (x=pm arccos(-a/2)+2kpi)`

with `abs(a/2) le 1`

Answer 2

`-1 <= -a/2 <= 1`

Explanation:

f(x) = sin x + asin 2x + sin 3x = 0
Apply the trig identity:
`sin a + sin b = 2sin ((a + b)/2)cos ((a - b)/2)`
In this case:
sin x + sin 3x = 2.sin 2x.cos x
f(x) = 2sin 2x.cos x + asin 2x = 0
f(x) = sin (2x)(2cos x + a) = 0
Either one of the 2 factors must be zero
2cos x + a = 0 --> `cos x = - a/2`
Condition:
`- 1 <= - a/2 <= 1`, or
I-a/2I <= 1