" (ii) If "A={x:(pi)/(6)<=x<=(pi)/(3)}" and "f(x)=cos x-x(1+x)

cos ^ (- 1) x + sin ^ (- 1) ((1) / (2) x) = (pi) / (6)

Solve : sin ^(-1)x - cos ^(-1) x = (pi )/(6)

tan(3x+(pi)/(6))=0

the range of cos (x+ (pi)/(6)) cos (x-(pi)/(6)) is

Maximum value of sin (x + (pi)/(6)) + cos (x + (pi)/(6)) is

lim_ (x rarr (pi) / (6)) (2 (sin x cos ((pi) / (6)) - cos x sin ((pi) / (6)))) / (x- (pi) / (6))

For what value of k is the following function continuous at x=-(pi)/(6)?f(x)={(sqrt(3)sin x+cos x)/(x+(pi)/(6)), when x!=-(pi)/(6)k, whenx =(pi)/(6)

The minimum value of the function tan(x+(pi)/(6))f(x)=(tan(x+(pi)/(6)))/(tan x) is:

if f(x)=3sin(x+(pi)/(6))+cos(x-(pi)/(6)) is always less then or equal to k then k is

find the range of function f(x)=sin(x+(pi)/(6))+cos(x-(pi)/(6))