document.write( "Question 945487: Solve the equation cotangent squared of the quantity of pi over 2 minus theta = secant of theta + 1 for for theta greater than or equal to 0 but less than 2 pi. \n" ); document.write( "
Algebra.Com's Answer #576690 by lwsshak3(11628)\"\" \"About 
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Solve the equation cotangent squared of the quantity of pi over 2 minus theta = secant of theta + 1 for for theta greater than or equal to 0 but less than 2 pi.
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\n" ); document.write( "use x for theta
\n" ); document.write( "\"cot%5E2%28pi%2F2-x%29=secx%2B1\"
\n" ); document.write( "\"tan%5E2%28x%29=1%2Fcosx%2B1\"
\n" ); document.write( "\"sin%5E2%28x%29%2Fcos%5E2%28x%29=1%2Fcosx%2B1\"
\n" ); document.write( "lcd: cos^2(x)
\n" ); document.write( "\"sin%5E2%28x%29=cosx%2Bcos%5E2%28x%29\"
\n" ); document.write( "\"1-cos%5E2%28x%29=cosx%2Bcos%5E2%28x%29\"
\n" ); document.write( "2cos^2(x)+cosx-1=0
\n" ); document.write( "(2cosx-1)(cosx+1)=0
\n" ); document.write( "cosx=1/2
\n" ); document.write( "x=π/3,5π/3
\n" ); document.write( "or
\n" ); document.write( "cosx=-1
\n" ); document.write( "x=π
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