Question

If \((\alpha,\beta)\subseteq\left(0,\dfrac{\pi}{2}\right)\) is the smallest set containing all the possible solutions of the inequality \[ \sin x+\cos2x>1, \] then \(\alpha+\beta=\)

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{\pi}{6}\)
• D. \(\dfrac{\pi}{4}\)

Hint:

Whenever an inequality involves both \(\sin x\) and \(\cos2x\), first convert \[ \boxed{\cos2x=1-2\sin^2x} \] so that the inequality becomes a quadratic in \(\sin x\).

Answer 1

The Correct Option is C

Step 1: Rewrite the inequality. Using \[ \cos2x=1-2\sin^2x, \] the given inequality becomes \[ \sin x+1-2\sin^2x>1. \] Hence, \[ \sin x-2\sin^2x>0. \] Factorizing, \[ \sin x(1-2\sin x)>0. \]

Step 2: Find the solution in \(\left(0,\dfrac{\pi}{2}\right)\). Since \[ 0<\sin x<1 \] in the interval \(\left(0,\dfrac{\pi}{2}\right)\), the inequality reduces to \[ 0<\sin x<\frac12. \] Therefore, \[ 0<x<\frac{\pi}{6}. \] Thus, \[ (\alpha,\beta)=\left(0,\frac{\pi}{6}\right). \]

Step 3: Find the required value. Hence, \[ \alpha+\beta = 0+\frac{\pi}{6} = \frac{\pi}{6}. \] Therefore, \[ \boxed{\frac{\pi}{6}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.