document.write( "Question 965359: sin(2x)sinx - cos(2x)cosx = 1 where x is (-pi, pi) \n" ); document.write( "
Algebra.Com's Answer #590148 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
sin(2x)sinx - cos(2x)cosx = 1 where x is (-pi, pi)
\n" ); document.write( "2sinxcosxsinx-(cos^2(x)-sin^2(x))cosx=1
\n" ); document.write( "2sin^2(x)-cos^2(x)+sin^2(x)=1
\n" ); document.write( "3sin^2(x)-cos^2(x)=1
\n" ); document.write( "3-3cos^2(x)-cos^2(x)=1
\n" ); document.write( "4cos^2(x)=2
\n" ); document.write( "cos^2(x)=2/4=1/2
\n" ); document.write( "cosx=±√(1/2=±√2/2
\n" ); document.write( "x=-3π/4, -π/4, π/4, 3π/4 \n" ); document.write( "
\n" );