Question

The number of values of x in the interval [0, 5$2\cos^{2} + 3\cos\theta - 2 = 0$] satisfying the equation $\cos\theta = \frac{- 3 \pm \sqrt{9 + 16}}{4} = \frac{- 3 \pm 5}{4}$is.

• A. 0
• B. 5
• C. 6
• D. 10

Answer 1

Answer: (C)

6

Answer 2

$\therefore$

$2x = \left( n + \frac{1}{2} \right)\ \pi\text{ or}(2n + 1)\ \pi$ $\Rightarrow$

$x = (2n + 1)\frac{\pi}{4}\text{ or }(2n + 1)\frac{\pi}{2}n = - 2, - 1,0,1,2$

$\therefore$ $( 3 \sin x - 1 ) ( \sin x - 2 ) = 0$ $\frac{- 3\pi}{2},\frac{- \pi}{2},\frac{\pi}{2}, ⥂ \frac{3\pi}{2},\frac{5\pi}{2}$ $- \pi \leq x \leq \pi$

$\Rightarrow$ $\sin 7\theta + \sin\theta - \sin 4\theta = 0$, ($\Rightarrow$)

Let $2\sin 4\theta\cos 3\theta - \sin 4\theta = 0$, $\Rightarrow$ are the solutions in $\sin 4\theta(2\cos 3\theta - 1) = 0 \Rightarrow \sin 4\theta = 0,\ \cos 3\theta = \frac{1}{2}$. Then $4\theta = 0 \Rightarrow 4\theta = \pi$ $\Rightarrow$ $\theta = \frac{\pi}{4}$, $\cos 3\theta = \frac{1}{2}$ are the

solutions in $\Rightarrow$.

$3\theta = \frac{\pi}{3}$ Required number of solutions = 6.