Possible values of $\theta$ for which the point $(\cos\theta, \sin\theta)$ lies inside the triangle formed by lines $x + y = 2; x - y = 1$ and $6x + 2y = \sqrt{10}$ are:
Step-by-Step Solution
Key Concept: Trigonometric expressions are converted to standard form using auxiliary angle method, then used to find intersection with geometric objects.
The line $6\cos\theta + 2\sin\theta = \sqrt{10}$ is rewritten using $6\cos\theta + 2\sin\theta = \sqrt{40}\sin(\theta + \alpha)$ where $\tan\alpha = 3$. From $\tan^{-1}3 = a$, we get $\sin(\theta + a) = \frac{1}{2}$, so $\theta + a = \frac{5\pi}{6}$, giving $\theta = \frac{5\pi}{6} - \tan^{-1}3$. The circle $x^2 + y^2 = 1$ intersects with lines $x - y = 1$, $x + y = 2$, and $6x + 2y = \sqrt{10}$.
Correct Answer: 1,2,3