Question 979386
<pre>
I think you're confusing sin(2x) with sin<sup>2</sup>(x)

You cannot treat trig functions like "sin" and "cos" as 
if they behaved like multication, because they don't!
To simplify trigonometric expressions and equations,
we us IDENTITIES:

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If the problem were as you solved it, you would
have had it half correct:

{{{sin^2(x) + sin(x)}}}{{{""=""}}}{{{"0"}}}

You would do as you did and factor out sin(x):

{{{sin(x)(sin(x)^""+1)}}}{{{""=""}}}{{{"0"}}}

{{{sin(x)=0}}},     {{{sin(x)+1=0}}}

{{{x=n*pi}}},      {{{sin(x)=-1}}}
                    {{{x=3pi/2+2pi*n}}}
                    {{{x=3pi/2+4pi*n/2}}}
                    {{{x=(3+4n)(pi/2)}}}

Your solution of the second part here was incorrect.

HOWEVER, THIS IS NOT THE EQUATION YOU ARE TO SOLVE!!!

Here is the problem:

{{{sin(2x) + sin(x)}}}{{{""=""}}}{{{"0"}}}

You cannot factor.

What you do is use the identity {{{sin(2theta)=2sin(theta)cos(theta)}}}

{{{2sin(x)cos(x)+sin(x)}}}{{{""=""}}}{{{"0"}}}

Now you can factor out sin(x)

{{{sin(x)(2cos(x)+1)=0}}}

{{{sin(x)}}},      {{{2cos(x)+1=0}}}
                   {{{2cos(x)=-1}}}
{{{x=n*pi}}},      {{{cos(x)=-1/2}}}
                    {{{x=2pi/3+2pi*n}}}   and   {{{x=4pi/3+2pi*n}}}
                    {{{x=2pi/3+6pi*n/3}}}       {{{x=4pi/3+6pi*n/3}}}
                    {{{x=(2+6n)(pi/3)}}}        {{{x=(4+6n)(pi/3)}}}
                    {{{x=2(1+3n)(pi/3)}}}       {{{x=2(2+3n)(pi/3)}}}
                    {{{x=(1+3n)(2pi/3)}}}       {{{x=(2+3n)(2pi/3)}}}

Edwin</pre>