document.write( "Question 252486: 2sin^2x+3cosx=3 \n" ); document.write( "
Algebra.Com's Answer #184428 by drk(1908)\"\" \"About 
You can put this solution on YOUR website!
Here is the original problem:
\n" ); document.write( "(i) 2sin^2x+3cosx=3
\n" ); document.write( "We have an identity for sin^2(x).
\n" ); document.write( "Sin^2(x) + Cos^2(x) = 1.
\n" ); document.write( "Solving for Sin^2(x), we get
\n" ); document.write( "sin^2(x) = 1 - cos^2(x)
\n" ); document.write( "Now, by substitution into (i), we get
\n" ); document.write( "2(1 - cos^2(x)) + 3cos(x) - 3 = 0
\n" ); document.write( "2cos^2(x) - 3cos(x) + 1 = 0.
\n" ); document.write( "Factoring, we get
\n" ); document.write( "(2cos(x) - 1)(cos(x) - 1) = 0.
\n" ); document.write( "Solve each parenthesis for x.
\n" ); document.write( "(2cos(x) - 1) = 0
\n" ); document.write( "cos(x) = 1/2.
\n" ); document.write( "x = 60 degrees or 300 degrees ; radians = pi/3, 5pi/3
\n" ); document.write( "(cos(x) - 1) = 0
\n" ); document.write( "cos(x) = 1
\n" ); document.write( "x = 0 degrees ; 0 radians.
\n" ); document.write( "-----
\n" ); document.write( "So, your answers are: 0, 60 degrees, 300 degrees. \n" ); document.write( "
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