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

2 Answers
Cesareo R.
Mar 21, 2017

See below.

Explanation:

Using and substituting

sin(3x)=3cos^2xsinx-sin^3xsin(3x)=3cos2xsinx−sin3x and
sin(2x)=2cosx sinxsin(2x)=2cosxsinx

3cos^2xsinx-sin^3x+sinx+2asinxcosx=03cos2xsinx−sin3x+sinx+2asinxcosx=0

or

(3cos^2x-sin^2+1+2acosx)sinx=0(3cos2x−sin2+1+2acosx)sinx=0

or

(4cos^2x+2acosx)sinx=0(4cos2x+2acosx)sinx=0

or

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

Nghi N.
Mar 22, 2017

-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

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