解x*sin(x)+2*cos(x)-2 = 0这个方程~谢谢我用maple只能得出> solve(x*sin(x)+2*cos(x)-2 = 0); 0, RootOf(4*sin(_Z)+sin(_Z)*_Z^2-4*_Z)就算用allvalues命令也算不出实数解有人能帮我算下么?要>0以后的五个解

已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x 2cos x (sin x -cos x)+1 y=2cos x (sin x+cos x) 化简sin(x)cos(x)cos(2x) 化简sin(x)cos(x)cos(2x) sin x cos x cos 2x 化简 (sin^2x+cos^2x )(sin^2x-cos^2x)怎么得sin^2x-cos^2x? 化简 cos^2(x)*sin^2(x)-sin^2(x) 化简 cos^(2)(x)*sin^(2)(x)-sin^(2)(x) 化简sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x 已知sin(x)+3cos(x)=2,求sin(x)-cos(x)/sin(x)+cos(x)=? sin(x/2)cos(x/2)= 化简sin^4x+cos^2x sin[cos^2(x^3+x)],求导 y=(sin x + COS x)^2 化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x)