document.write( "Question 901911: Solve the trigonometric inequality. Write the solution in interval notation.\r
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document.write( "2cos²(3x) + 5cos(3x) - 3 \n" );
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Algebra.Com's Answer #547025 by Edwin McCravy(20065) You can put this solution on YOUR website! \r\n" ); document.write( "2cos²(3x) + 5cos(3x) - 3 < 0\r\n" ); document.write( "\r\n" ); document.write( "The left factors just as 2u²+5u-3 = (u+3)(2u-1)\r\n" ); document.write( "\r\n" ); document.write( "[cos(3x)+3][2cos(3x)-1] < 0\r\n" ); document.write( "\r\n" ); document.write( "We find the critical values by finding the zeros of the left side\r\n" ); document.write( "\r\n" ); document.write( "The first factor has no zeros since no cosine can be -3\r\n" ); document.write( "The second factor has zeros when cos(3x) =\n" ); document.write( " |