[2cos^2(x/2)]/[2cosx(-1/2-cos2x)]/[cosx(-1/2-cos2x)]-sinx-1=4cos^2(x/2)/[cos^2x(1+2cos2x)^2]-sinx-1cos2x=2cos^2x-1tan2x=2tanx/(1-tan^2x)tanx=√2且x属于第I象限,不会在第三象限.因为tan2x