Use calculus to prove the inequality, `sin x >= (2x)/pi` in `0 <= x <= pi/2.`

Number of integrall solutions of the inequality 2sin^(2)x-5sin x+2>0 in x in[0,2 pi] is

Use this inequality to prove that,cos x<=1-(x^(2))/(pi) in 0<=x<=(pi)/(2)

The solution set of x in(-pi,pi) for the inequality sin 2x+1 le cosx+2sin x is

The solution set of x in(-pi,pi) for the inequality sin 2x+1 le cosx+2sin x is

Find the solution set of the inequality 4sin^(2)x-8sin x+3<=0,0<=x<=2 pi

Using calculus establish the inequality, (x^(b)+y^(b))^(1//b) lt ( x^(a)+y^(a))^(1//a) , where x gt 0 , y gt 0 and b gt a gt 0 .

The solution set of x in(-pi,pi) for the inequality sin2x+1<=cos x+2sin x is: x in[0,(pi)/(6)]( b) x in[(pi)/(6),(5 pi)/(6)]uu{0}x in[-(pi)/(6),(6 pi)/(6)] (d) None of these