Trigonometric Equations
Trig Equations Inequations
nta_abhyas_2025
Grade 11
Question:
If $\int_{0}^{1} \sin\theta$ and $\tan\theta$ are in G.P., then the complete solution set of $\theta$ is
\{\theta : \theta = 2n\pi + \left(\frac{\pi}{3}\right), n \in I\}
\{\theta : \theta = 2n\pi + \left(\frac{\pi}{3}\right), n \in I\}
\{\theta : \theta = n\pi + (-1)^n\left(\frac{\pi}{3}\right), n \in I\}
\{\theta : \theta = n\pi + \frac{\pi}{3}, n \in I\}
Step-by-Step Solution
Key Concept: The maximum of $a\sin x + b\cos x$ is $\sqrt{a^2 + b^2}$, and matching LHS = RHS requires finding where both sides equal their extremal values.
We have $\text{LHS} = 12\sin x + 5\cos x = \sqrt{12^2 + 5^2}\sin(x + \alpha) = \sqrt{144 + 25}\sin(x + \alpha) = 13\sin(x + \alpha)$ where $\tan\alpha = \frac{5}{12}$. The maximum value is 13 when $\sin(x + \alpha) = 1$. The RHS equals $2(t^2 - 4t + 4) + 13 = 2(t - 2)^2 + 13 \geq 13$. For the equation LHS = RHS to have solutions, we need $13\sin(x + \alpha) = 13$, which gives $\sin(x + \alpha) = 1$ and $t = 2$. This yields $12\sin x + 5\cos x = 13$, so $\sin(x + \alpha) = 1$.
Correct Answer: 13