Question 1019974
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I need help on solving this trig equation:
(sin2x)(cosx)+(cos2x)(sinx)=(-1/2)
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<pre>
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
<A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A> in this site).

Take {{{alpha}}} = 2x, {{{beta}}} = x.

Then you get your equation in this form

sin(3x) = {{{-1/2}}}.

The solutions are 

3x =  {{{7pi/6}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . .  and   3x = {{{11pi/6}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . . 


In terms of x the solutions are 

x = {{{7pi/18}}} + {{{(2/3)*k*pi}}}, k = 0, +/-1, +/-2, . . .  and   x = {{{11pi/18}}} + {{{(2/3)*k*pi}}}, k = 0, +/-1, +/-2, . . . 
</pre>