document.write( "Question 1098062: Solve given that 0 ≤ x < 2π\r
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document.write( "--Not sure what to select for the topic of this question, and I am not sure what the process is to solve it. I would appreciate any pointers, thank you! \n" );
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Algebra.Com's Answer #803196 by CubeyThePenguin(3113) You can put this solution on YOUR website! cos(2x) = 1 - 2sin^2(x)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Rewrite as a quadratic in terms of sin(x). Let sin(x) = y if you like. \n" ); document.write( "1 - 2sin^2(x) - sin(x) = 0 \n" ); document.write( "-2y^2 - y + 1 = 0 \n" ); document.write( "2y^2 + y - 1 = 0 \n" ); document.write( "(2y - 1)(y + 1) = 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "sin(x) = 1/2 --> x = pi/6 \n" ); document.write( "sin(x) = -1 --> x = 3pi/2 \n" ); document.write( " |