document.write( "Question 312174: Please help me with this question and show all steps if possible thank you!
\n" ); document.write( "\"3+cos2x=+cosx+%2B2\"
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Algebra.Com's Answer #223191 by fractalier(6550)\"\" \"About 
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Okay from
\n" ); document.write( "3 cos2x = cosx +2
\n" ); document.write( "let's apply our identity for cos 2x...
\n" ); document.write( "cos 2x = cos^2 x - sin^2 x = 2 cos^2 x - 1
\n" ); document.write( "so that if we substitute we get
\n" ); document.write( "3(2 cos^2 x - 1) = cos x + 2
\n" ); document.write( "6 cos^2 x - 3 = cos x + 2
\n" ); document.write( "6 cos^2 x - cos x - 5 = 0 now factor this...
\n" ); document.write( "(6 cos x + 5)(cos x - 1) = 0
\n" ); document.write( "and then
\n" ); document.write( "cos x = -5/6 or cos x = 1
\n" ); document.write( "x = arc cos (-5/6) or x = 0
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