<PRE><B>Question 1201169</B>
&lt;br&gt;
We don&#39;t know what the student&#39;s error was, because we don&#39;t know HOW he got his answer.&lt;br&gt;
Both of the responses you have received show some poor algebraic practices; in both responses, both sides of an equation are divided by cos(x), but cos(x) might be 0.&lt;br&gt;
{{{sin(2x)/cos(x)=2}}}
{{{sin(2x)=2cos(x)}}}&lt;br&gt;
Notice there I multiplied both sides of the equation by cos(x); since cos(x) can be 0, I will have to check any solutions I come up with.&lt;br&gt;
{{{2sin(x)cos(x)=2cos(x)}}}
{{{2sin(x)cos(x)-2cos(x)=0}}}
{{{2cos(x)(sin(x)-1)=0}}}
{{{cos(x)=0}}}  or  {{{sin(x)=1}}}&lt;br&gt;
We know cos(x) can&#39;t be 0, so the only possibility is that sin(x)=1.  But when sin(x)=1, cos(x)=0.&lt;br&gt;
So there are no solutions to the equation in ANY interval.&lt;br&gt;
Correct ANSWER: No solutions&lt;br&gt;
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