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

To solve the given equation \( (4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}} \), we need to find the number of solutions in the interval \( x \in \left[-2\pi, \frac{5\pi}{2}\right] \).

1. First, simplify the right-hand side of the equation: 

\[\frac{-4}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{-4(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{-4 + 4\sqrt{3}}{-2} = 2 - 2\sqrt{3}\]

1.  So the equation becomes: 

\[(4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = 2 - 2\sqrt{3}\]

1. Using the identity \(\cos^2 x = 1 - \sin^2 x\), rewrite the equation: 

\[(4 - \sqrt{3}) \sin x - 2\sqrt{3} (1 - \sin^2 x) = 2 - 2\sqrt{3}\]

1.  Simplify it to: 

\[(4 - \sqrt{3}) \sin x - 2\sqrt{3} + 2\sqrt{3} \sin^2 x = 2 - 2\sqrt{3}\]

1. Move all terms involving \(\sin x\) and constants to one side: 

\[2\sqrt{3} \sin^2 x + (4 - \sqrt{3}) \sin x = 0\]

1.  Factor out \(\sin x\): 

\[\sin x (2\sqrt{3} \sin x + 4 - \sqrt{3}) = 0\]

1.  This gives two separate equations to solve:
   • \(\sin x = 0\)
   • \(2\sqrt{3} \sin x + 4 - \sqrt{3} = 0\)
2. For \(\sin x = 0\), the solutions within the interval \([−2\pi, \frac{5\pi}{2}]\) are:
   • \(x = -2\pi, 0, \pi, 2\pi\)
3. For the equation \(2\sqrt{3} \sin x = \sqrt{3} - 4\), simplify: 

\[\sin x = \frac{\sqrt{3} - 4}{2\sqrt{3}}\]

1.  Calculate: 

\[\approx -0.633\]

1.  Find where this value occurs within the given domain.
2. Solutions for \(\sin x = -0.633\) exist at specific reference angles computed via the inverse sine function, yielding one solution in each relevant period.
3. Combining all found solutions:
   • Four solutions from \(\sin x = 0\): \(x = -2\pi, 0, \pi, 2\pi\).
   • Additional solution with the reference angle for \(\sin x = -0.633\) within \([-2\pi, \frac{5\pi}{2}]\).
4. Thus, the number of solutions is \(\textbf{5}\).

Answer 2

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.