Question

Sol. [Key Concept: Put $(\cos\ \theta,\sin\ \theta)$ inside all three lines; the binding one gives $\theta in(0,\ \frac{5\pi}{6} - \tan^{- 1}3)$.]

For Q:

$${6cos\theta + 2sin\theta = \sqrt{10} }{\tan^{- 1}3 = \alpha,sin(\theta + \alpha) = \frac{1}{2} }$$

$$\theta + \alpha = \frac{5\pi}{6}$$

$$\theta = \frac{5\pi}{6} - \tan^{- 1}3$$

$$\therefore\ \theta \in \left( 0,\frac{5\pi}{6} - \tan^{- 1}3 \right)$$

✔ (a) $\frac{\pi}{8}$ lies in $(0,\ \frac{5\pi}{6} - \tan^{- 1}3)$.

✔ (b) $\frac{\pi}{4}$ lies in the interval.

✔ (c) $\frac{3\pi}{8}$ lies in the interval.

✘ (d) $\frac{\pi}{2}$ exceeds $\frac{5\pi}{6} - \tan^{- 1}3 \approx {78.4}^{\circ}$.