SOLUTION: I need help on solving this trig equation: (sin2x)(cosx)+(cos2x)(sinx)=(-1/2)

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Question 1019974: I need help on solving this trig equation:
(sin2x)(cosx)+(cos2x)(sinx)=(-1/2)
Found 2 solutions by ikleyn, KMST:
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
I need help on solving this trig equation:
(sin2x)(cosx)+(cos2x)(sinx)=(-1/2)
---------------------------------------- 

Apply the addition formula for sine:

sin(alpha+beta) = sin(alpha)*cos(beta) + cos(alpha)*sin(beta).

(This is one of the basis formulas in Trigonometry.  See the lesson
Addition and subtraction formulas in this site).

Take  = 2x,  = x.

Then you get your equation in this form

sin(3x) = .

The solutions are 

3x =   + , k = 0, +/-1, +/-2, . . .  and   3x =  + , k = 0, +/-1, +/-2, . . . 


In terms of x the solutions are 

x =  + , k = 0, +/-1, +/-2, . . .  and   x =  + , k = 0, +/-1, +/-2, . . .


Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
The left side of that equation should remind you of the trigonometric identity
.
Making (or the other way around),
, or
.
So, we can re-write the equation as
.
In the interval, the solutions for are
or .
We could write that as .
Since sine has a period of ,
other values for differing in any multiple of also have .
All the solutions can be written as
for any integer .
So, for any integer , or
for any integer .
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