document.write( "Question 391582: How do you solve for x in cos2x = cosx, -pi<=x<=pi? \n" ); document.write( "

Algebra.Com's Answer #277868 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"cos+2x+=+cos+x\"$ <==> $\"2%28cos+x%29%5E2+-+1+=+cosx\"$ <==> $\"2%28cos+x%29%5E2+-+cosx+-+1+=+0\"$ <==> (2 cosx + 1)(cosx - 1) = 0 ==> cosx = -1/2 or cos x = 1.
\n" ); document.write( "From cosx = -1/2, x = $\"-2pi%2F3\"$ , $\"+2pi%2F3\"$ .
\n" ); document.write( "From cos x = 1, x = 0.
\n" ); document.write( "The solutions are then 0, $\"-2pi%2F3\"$ , $\"+2pi%2F3\"$ . \n" ); document.write( "

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