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OpenStudy (anonymous):
sin x cos 2x + cos x sin 2x = 1/2 between 0<= x< 360 I got up to sinx(2cos^2-1)+cosx(2cosxsinx)=1/2 2cos^2xsinx-sinx+2cos^2xsinx=1/2 4cos^2xsinx-sinx=1/2 sinx(4cos^2x-1)=1/2 2sinx(4cos^2x-1)=1
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OpenStudy (anonymous):
well I did go more forward by distributing (8cos^2xsinx-2sinx)=1 sin2x(4cosx)-2sinx=1 (But I dont think this is right though)
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