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

2 Answers
Cesareo R.
Mar 21, 2017

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#

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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