document.write( "Question 1206254: I need help figuring out all the solutions in the interval [0,2pi)
\n" ); document.write( "sin^2(x)+2cos(x)=2 \n" ); document.write( "

Algebra.Com's Answer #843546 by ikleyn(52786) $\"\"$ $\"About$

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\n" ); document.write( "I need help figuring out all the solutions in the interval [0,2pi)
\n" ); document.write( "sin^2(x)+2cos(x)=2
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\n" ); document.write( "\n" ); document.write( " The idea is to replace sin^2(x) by expression which contains cos(x), only.\r
\n" ); document.write( "\n" ); document.write( " Then the entire equation will be relative cos(x), and we will be able solve it.\r
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document.write( "So, by implementing this idea, we write  sin^2(x) = 1 - cos^2(x).\r\n" );
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document.write( "Then the given equation takes the form\r\n" );
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document.write( "    (1-cos^2(x)) + 2cos(x) = 2.\r\n" );
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document.write( "Simplify it\r\n" );
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document.write( "    - cos^2(x) + 2cos(x) = 1\r\n" );
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document.write( "    cos^2(x) - 2cos(x) + 1 = 0\r\n" );
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document.write( "     = 0   --->\r\n" );
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document.write( "    cos(x) = 1  --->  x = 0.\r\n" );
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document.write( "ANSWER.  In the given interval, the original equation has a UNIQUE root  x = 0 radians.\r\n" );
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