document.write( "Question 459274: Find to the nearest tenth of a degree ,all values of x in the interval 0 less than/equal to x lessthan/equal to 360 degrees that satisfy the equation 4cos^2x-5sinx-5=0\r
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Algebra.Com's Answer #314972 by Gogonati(855)\"\" \"About 
You can put this solution on YOUR website!
First we simplify this equation:\"4%281-%28sinx%29%5E2%29-5sinx-5=0\"=>\r
\n" ); document.write( "\n" ); document.write( "=>\"4-4%28sinx%29%5E2-5sinx-5=0\"=>\"4%28sinx%29%5E2%2B5sinx%2B1=0\". \r
\n" ); document.write( "\n" ); document.write( "In the last equation we substitute \"sinx=y\", and get the quadratic equation:\r
\n" ); document.write( "\n" ); document.write( "\"4y%5E2%2B5y%2B1=0\", solving this equation we get: y=-1 and y=-1/4.\r
\n" ); document.write( "\n" ); document.write( "Remembering our substitution \"y=+sinx\" we need to solve two simple \r
\n" ); document.write( "\n" ); document.write( "trigonometric equations equivalent with our original equation.\r
\n" ); document.write( "\n" ); document.write( "1)\"sinx=-1\", where \"x=3%2Api%2F2\", and\r
\n" ); document.write( "\n" ); document.write( "2)\"sinx=-1%2F4\", where \"x=-0.25\", and \"x=2%2Api%2B0.25\".
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