If `|cos x + sin x| + | cos x - sin x| = 2 sin x; x in [0,pi]`, then maximum integral value of x is :

The maximum value of 6 sin x cos x is

The maximum value of sin x cos x is ……..

(1 + sin 2x + cos 2x )/( cos x + sin x ) = 2 cos x

Solve cos^(-1) (cos x) gt sin^(-1) (sin x), x in [0, 2pi]

Statement I The system of linear equations x+(sin alpha )y+(cos alpha )z=0 x+(cos alpha ) y+(sin alpha )z=0 -x+(sin alpha )y-(cos alpha )z=0 has a not trivial solution for only one value of alpha lying between 0 and pi . Statement II |{:(sin x, cos x , cos x),( cos x , sin x , cos x) , (cos x , cos x , sin x ):}|=0 has no solution in the interval -pi//4 lt x lt pi//4 .

if sin x + sin^2 x = 1 , then the value of cos^2 x + cos^4x is

if sin x + sin^2 x = 1 , then the value of cos^2 x + cos^4x is

int_(0)^(pi//2) ( cos x - sin x )/( 2 + sin x cos x ) dx is equal to

If |sin x+cos x|=|sin x|+|cos x| (sin x ,cos x !=0) , then in which quadrant does x lie?

Prove that: sin 3x + sin 2x - sin x = 4 sin x cos x/2 cos (3x)/2