document.write( "Question 596831: solve \"2+cos%5E2%28x%29+-+5cos%28x%29+%2B+2+=+0\" for all principal values of x.\r
\n" ); document.write( "\n" ); document.write( "I am not sure where nor how to proceed with problem. Am I supposed to use a Pythagorean identity, and if so, where do I go from there. Any help you can give me would be so greatly appreciated. I am struggling through my precalculus book, and I did not do well in Trig or Geometry and passes Algebra I and II with a mid C. Please help me understand this problem.
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Algebra.Com's Answer #377910 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "First of all, I fixed your equation. means the cosine of the angle , whereas is the squared value of the cosine function at . Very different things and they didn't give you\r
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\n" ); document.write( "\n" ); document.write( "in a Precalculus book. \r
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\n" ); document.write( "\n" ); document.write( "The easiest way to see what is going on in this equation is to use a simple substitution.\r
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\n" ); document.write( "\n" ); document.write( "Let and substitute.\r
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\n" ); document.write( "\n" ); document.write( "Factor the quadratic in \r
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\n" ); document.write( "\n" ); document.write( "Apply the Zero Product Rule:\r
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\n" ); document.write( "\n" ); document.write( "Now recall the RANGE of the cosine function, i.e. . That means that is an extraneous root and can be discarded.\r
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\n" ); document.write( "\n" ); document.write( "Hence, substituting back\r
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\n" ); document.write( "\n" ); document.write( "The principal values of the argument for cosine are in the interval\r
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\n" ); document.write( "\n" ); document.write( "Find the two angles in the unit circle where the -coordinate is equal to and select the one that is in the principal value interval.\r
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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