Question

The number of solutions of the given equation$3\pi - \alpha,$ where $4\pi + \alpha$is.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

$- 1 \leq y \leq 1 \Rightarrow - 1 \leq 1 - \sqrt{3 + 2\alpha} \leq 1$ ….(i)

Also we have

$\Rightarrow$ …..(ii)

$3 + 2\alpha \leq 4 \Rightarrow \alpha \leq \frac{1}{2} - \frac{3}{2} \leq \alpha \leq \frac{1}{2}$ …..(iii)

Now (i) and (iii) gives,

$\tan \theta = \frac { 1 } { 2 } \left( \sqrt { 3 } - \frac { 1 } { \sqrt { 3 } } \right) = \frac { 1 } { \sqrt { 3 } } = \tan \left( \frac { \pi } { 6 } \right)$ $\cos\alpha = \frac{3}{5}$ $\sin\alpha = \frac{4}{5}$.

$3\cos\theta + 4\sin\theta = k$ Solutions for $5\cos(\theta - \alpha) = k \Rightarrow \cos(\theta - \alpha) = \pm 1$ are $\Rightarrow$ and $\theta - \alpha = 0^{ο},180^{o} \Rightarrow \theta = \alpha,\ 180^{o} + \alpha$.

Hence there are two solutions.