Question

The number of solutions of the equation \( (4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}} \), \( x \in \left[-2\pi, \frac{5\pi}{2}\right] \) is

• A. 4
• B. 3
• C. 6
• D. 5

Hint:

Convert the given trigonometric equation into a quadratic equation in \( \sin x \) or \( \cos x \). Solve the quadratic equation to find the possible values of the trigonometric function. Then, find the number of solutions for these values within the specified interval. Pay careful attention to the boundaries of the interval.

Answer 1

The Correct Option is D

The given equation is: \[ (4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}} \] Rationalize the right-hand side: \[ \frac{-4}{1 + \sqrt{3}} = \frac{-4(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{-4(1 - \sqrt{3})}{1 - 3} = \frac{-4(1 - \sqrt{3})}{-2} = 2(1 - \sqrt{3}) = 2 - 2\sqrt{3} \] Substitute \( \cos^2 x = 1 - \sin^2 x \) into the equation: \[ (4 - \sqrt{3}) \sin x - 2\sqrt{3} (1 - \sin^2 x) = 2 - 2\sqrt{3} \] \[ (4 - \sqrt{3}) \sin x - 2\sqrt{3} + 2\sqrt{3} \sin^2 x = 2 - 2\sqrt{3} \] \[ 2\sqrt{3} \sin^2 x + (4 - \sqrt{3}) \sin x - 2 = 0 \] This is a quadratic equation in \( \sin x \). Let \( y = \sin x \): \[ 2\sqrt{3} y^2 + (4 - \sqrt{3}) y - 2 = 0 \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ y = \frac{-(4 - \sqrt{3}) \pm \sqrt{(4 - \sqrt{3})^2 - 4(2\sqrt{3})(-2)}}{2(2\sqrt{3})} \] \[ y = \frac{-4 + \sqrt{3} \pm \sqrt{16 - 8\sqrt{3} + 3 + 16\sqrt{3}}}{4\sqrt{3}} \] \[ y = \frac{-4 + \sqrt{3} \pm \sqrt{19 + 8\sqrt{3}}}{4\sqrt{3}} \] Note that \( 19 + 8\sqrt{3} = 16 + 3 + 2 \cdot 4 \cdot \sqrt{3} = (4 + \sqrt{3})^2 \). \[ y = \frac{-4 + \sqrt{3} \pm (4 + \sqrt{3})}{4\sqrt{3}} \] Two possible values for \( y = \sin x \): \[ \sin x = \frac{-4 + \sqrt{3} + 4 + \sqrt{3}}{4\sqrt{3}} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2} \] \[ \sin x = \frac{-4 + \sqrt{3} - 4 - \sqrt{3}}{4\sqrt{3}} = \frac{-8}{4\sqrt{3}} = \frac{-2}{\sqrt{3}} \] Since \( -1 \le \sin x \le 1 \), \( \sin x = \frac{-2}{\sqrt{3}} \) is not possible. So, we need to find the number of solutions for \( \sin x = \frac{1}{2} \) in the interval \( \left[-2\pi, \frac{5\pi}{2}\right] \). The general solutions are \( x = n\pi + (-1)^n \frac{\pi}{6} \). For \( n = -2 \): \( x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6} \) For \( n = -1 \): \( x = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6} \) For \( n = 0 \): \( x = \frac{\pi}{6} \) For \( n = 1 \): \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) For \( n = 2 \): \( x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \) For \( n = 3 \): \( x = 3\pi - \frac{\pi}{6} = \frac{17\pi}{6}>\frac{15\pi}{6} = \frac{5\pi}{2} \) The solutions in the given interval are \( -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6} \). There are 5 solutions.