Animated Solution for Mathematics - Trigonometry: All possible values of $\theta \in [0,2\pi ]$ for which $\sin 2\theta +\tan 2\theta >0$ lie in:

We are tasked with solving the trigonometric inequality $\sin 2\theta +\tan 2\theta >0$ for the domain $\theta \in [0,2\pi ]$. While the expression appears straightforward, it requires a systematic approach to identify the valid intervals for $\theta$.

To simplify the expression, we express $\tan 2\theta$ in terms of sine and cosine. Substituting $\tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta }$, the inequality becomes:

By factoring out $\sin 2\theta$, we transform the sum into a product:

Combining the terms inside the parentheses leads to:

Recognizing that $\frac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta$, we arrive at the simplified form:

Consider the term $(1+\cos 2\theta )$. Since the range of $\cos 2\theta$ is $[-1,1]$, the expression $(1+\cos 2\theta )$ is always non-negative, falling within the interval $[0,2]$.

Because this term is never negative, it does not affect the sign of the inequality. However, it acts as a constraint: for the product to be strictly greater than zero, we must ensure $1 + \cos 2\theta eq 0$. If $1+\cos 2\theta =0$, the entire expression becomes zero, violating the strict inequality.

Given $\theta \in [0,2\pi ]$, the argument $2\theta$ spans the range $[0,4\pi ]$. We require $\tan 2\theta >0$, which occurs in the first and third quadrants.

In the range $[0,4\pi ]$, this condition is satisfied when:

We must also satisfy the constraint $\cos 2\theta eq -1$. This occurs when $2\theta =\pi ,3\pi ,5\pi ,\dots$. Since these values are already excluded by our open intervals, the condition is naturally satisfied.

To determine the final set of values for $\theta$, we divide each interval by $2$. This yields the solution set:

By factoring, identifying the non-negative term, and carefully mapping the domain, we have successfully solved the inequality. This systematic rigor is the key to mastering JEE-level trigonometric challenges.