Given `A=sin^2theta+cos^4theta,` then for all real `theta,` (a)`1lt=Alt=2` (b) `3/4lt=Alt=1` (c)`(13)/(16)lt=Alt=1` (d) `3/4lt=Alt=(13)/(16)`

Given A=sin^(2)theta+cos^(4)theta, then for all real theta(a)1<=A<=2(b)(3)/(4)<=A<=1(c)(13)/(16)<=A<=1(d)(3)/(4)<=A<=(13)/(16)

If A""=""sin^2x""+""cos^4x , then for all real x : (1) 3/4lt=Alt=1 (2) (13)/(16)lt=Alt=1 (3) 1lt=Alt=2 (4) 3/4lt=Alt=(13)/(16)

If A=cos^(2)theta+sin^(4)theta then for all values of theta(A)1<=A<=2(B)(13)/(16)<=A<=1(C)(3)/(4)<=A<=(13)/(16) (D) (3)/(4)<=A<=1

If A=cos^2theta+sin^4theta, prove that 3/4lt=Alt=1 for all values of thetadot

If cos theta cos 2 theta cos 3theta = 1//4 for 0 lt theta lt pi then theta =

If A = sin^2 theta+ cos^4 theta , then for all real values of theta .............. A) y in [1,2] B) y in [13//16,1] C) y in [3//4, 13//16] D) y in [3//4, 1]

Solve: sin 7 theta + sin 4 theta + sin theta = 0, 0 lt theta lt ( pi)/(2)

The solution of costheta.cos2theta.cos3theta=1/4,0 lt theta lt pi/4 is:

If 0 lt theta lt pi/2 and sin 2 theta = cos 3 theta, then the value of sin theta is-