document.write( "Question 594162: Solve the equation 2sin^2A = cosA + 2 over the interval [0,2π] \n" ); document.write( "
Algebra.Com's Answer #376807 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Solve the equation 2sin^2A = cosA + 2 over the interval [0,2π] \n" ); document.write( "2sin^2=cos+2 \n" ); document.write( "2(1-cos^2)=cos+2 \n" ); document.write( "2-2cos^2=cos+2 \n" ); document.write( "2cos^2+cos=0 \n" ); document.write( "cos(2cos+1)=0 \n" ); document.write( ".. \n" ); document.write( "cosA=0 \n" ); document.write( "A=π/2 and 3π/2 \n" ); document.write( ".. \n" ); document.write( "2cosA+1=0 \n" ); document.write( "cosA=-1/2 \n" ); document.write( "A=2π/3 and 5π/3 (in quadrants III and IV where cos<0) \n" ); document.write( " |