Question

Given that 0<x< 2π, solve the followi ing inequalities. (2) 2sin2x <3cosx (3) 2cosx3 sinx+10

Answer 1

2 cos^2 x + sin x -1 > 0 = > 2 -2 sin^2 x + sin x -1 > 0 2 sin^2 x - sin x -1 < 0 = > (sin x -1) (sin x + 1/2) < 0 sin x + 1/2 < 0 & sin x -1 > 0 or sin x -1 < 0 & sin x + 1/2 > 0 sin x < 1 & sin x > -1/2 x epsilon [0, 7 pi/6] 2 sin^2 x < 3 cos x = > 2 -2 cos^2x < 3 cos x 2 cos^2 x = 3 cos x -2 > 0 = > (cos x -1/2) (cos x + 2) > 0 cos x < 1/2 & cos x < -2 or cos x > 1/2 & cos x > -2 = > x elementof [0, pi/3) 2 cos^2 x + squareroot 3 sin x + 1 > 0 = > 2 -2 sin^2 x + squareroot 3 sin x