Question

All possible values of θ ∈ [0, 2π] for which sin 2θ + tan 2θ>0 lie in :

• A. (0, π/2) ∪ (π, 3π/2)
• B. (0, π/4) ∪ (π/2, 3π/4) ∪ (3π/2, 11π/6)
• C. (0, π/2) ∪ (π/2, 3π/4) ∪ (π, 7π/6)
• D. (0, π/4) ∪ (π/2, 3π/4) ∪ (π, 5π/4) ∪ (3π/2, 7π/4)

Hint:

When solving $\tan(n\theta)>0$, the intervals will be $1/n$ times the length of the standard $\tan \theta$ intervals.

Answer 1

The Correct Option is D

Step 1: $\sin 2\theta(1 + \frac{1}{\cos 2\theta})>0 \Rightarrow \sin 2\theta \frac{(\cos 2\theta + 1)}{\cos 2\theta}>0$.
Step 2: Since $\cos 2\theta + 1>0$ always (except where $\cos 2\theta = -1$), we need $\frac{\sin 2\theta}{\cos 2\theta}>0 \Rightarrow \tan 2\theta>0$.
Step 3: $\tan 2\theta>0$ when $2\theta \in (0, \pi/2) \cup (\pi, 3\pi/2) .......$
Step 4: Dividing by 2 and applying $2\pi$ periodicity gives the intervals in (D).