document.write( "Question 371832: Solve the equation over the interval [0 pi, 2 pi) : 3tan^3x=tanx \n" ); document.write( "

Algebra.Com's Answer #264860 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"3%28tanx%29%5E3+-+tanx+=+0\"$ ,
\n" ); document.write( " $\"tanx%283%28tanx%29%5E2+-+1%29+=+0\"$ ,
\n" ); document.write( "tanx = 0, or $\"%28tanx%29%5E2+=+1%2F3\"$ , or
\n" ); document.write( "tanx = 0, or $\"tanx+=+1%2Fsqrt%283%29\"$ or $\"tanx+=+-1%2Fsqrt%283%29\"$ .
\n" ); document.write( " therefore
\n" ); document.write( "x = 0, $\"pi\"$ , $\"pi%2F6\"$ , $\"7pi%2F6\"$ , $\"5pi%2F6\"$ , $\"11pi%2F6\"$ . \n" ); document.write( "

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