document.write( "Question 148293: Hi,\r
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document.write( "Question is solve the equation on the interval [0,2Pi0.\r
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document.write( "2cos^2x+sinx-2=0\r
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document.write( "Since cos^2x= 1-sin^2x I substituded this:\r
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document.write( "2(1-sin^2x) + sinx-2=0 to get
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document.write( "-2sin^2x + sinx=0 since -2 and +2 cancel eachother out. I am lost after this and the calculator hmm don't know whats worse figuring out the calculator or the math problem. \n" );
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Algebra.Com's Answer #108694 by stanbon(75887) ![]() You can put this solution on YOUR website! Question is solve the equation on the interval [0,2Pi]. \n" ); document.write( "2cos^2x+sinx-2=0 \n" ); document.write( "Since cos^2x= 1-sin^2x I substituded this: \n" ); document.write( "2(1-sin^2x) + sinx-2=0 to get \n" ); document.write( "-2sin^2x + sinx=0 since -2 and +2 cancel each other out. \n" ); document.write( "2sin^2x - sinx = 0 \n" ); document.write( "Now factor to get: \n" ); document.write( "sinx(2sinx-1) = 0 \n" ); document.write( "sinx = 0 or 2sinx = 1 \n" ); document.write( "sinx = 0 or sinx = 1/2 \n" ); document.write( "x = 0,pi,2pi or x = pi/6,(5/6)pi \n" ); document.write( "=============== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |