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

Question

Prepp Team · Accepted Answer

Simplify Trigonometric Equation

First, simplify the constant term on the right side of the equation:

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

Substitute $ \cos^2 x = 1 - \sin^2 x $ into the original equation:

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

Rearrange the terms to form a quadratic equation in $ \sin x $:

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

Solve for sin x

Let $ y = \sin x $. The equation becomes $ 2\sqrt{3} y^2 + (4-\sqrt{3}) y - 2 = 0 $. Apply the quadratic formula $ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 2\sqrt{3} $, $ b = 4-\sqrt{3} $, and $ c = -2 $.

Calculate the discriminant $ \Delta $:

\[\Delta = (4-\sqrt{3})^2 - 4(2\sqrt{3})(-2) = (16 - 8\sqrt{3} + 3) + 16\sqrt{3} = 19 + 8\sqrt{3}\]

Simplify the square root of the discriminant:

\[\sqrt{\Delta} = \sqrt{19 + 8\sqrt{3}} = \sqrt{16 + 3 + 2\sqrt{16 \times 3}} = \sqrt{(\sqrt{16} + \sqrt{3})^2} = 4 + \sqrt{3}\]

Find the possible values for $ y $:

\[y = \frac{-(4-\sqrt{3}) \pm (4+\sqrt{3})}{4\sqrt{3}}\]

This gives two possibilities:

$ y_1 = \frac{-4+\sqrt{3} + 4+\sqrt{3}}{4\sqrt{3}} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2} $
$ y_2 = \frac{-4+\sqrt{3} - (4+\sqrt{3})}{4\sqrt{3}} = \frac{-8}{4\sqrt{3}} = -\frac{2}{\sqrt{3}} $

The value $ y_2 = -\frac{2}{\sqrt{3}} $ is less than -1, so it is not a possible value for $ \sin x $. Thus, we only consider $ \sin x = \frac{1}{2} $.

Determine Solutions in Interval

The general solution for $ \sin x = \frac{1}{2} $ is $ x = n\pi + (-1)^n \frac{\pi}{6} $, where $ n $ is an integer.

We need to find the solutions within the interval $ x \in [-2\pi, \frac{5\pi}{2}] $.

Evaluate the general solution for different integer values of $ n $:

$ n = 0 \implies x = \frac{\pi}{6} $
$ n = 1 \implies x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} $
$ n = 2 \implies x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} $
$ n = -1 \implies x = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6} $
$ n = -2 \implies x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6} $

Verify which of these solutions fall within $ [-2\pi, \frac{5\pi}{2}] \approx [-2\pi, 2.5\pi] $:

$ \frac{\pi}{6} \in [-2\pi, \frac{5\pi}{2}] $
$ \frac{5\pi}{6} \in [-2\pi, \frac{5\pi}{2}] $
$ \frac{13\pi}{6} \approx 2.17\pi \in [-2\pi, \frac{5\pi}{2}] $
$ -\frac{7\pi}{6} \approx -1.17\pi \in [-2\pi, \frac{5\pi}{2}] $
$ -\frac{11\pi}{6} \approx -1.83\pi \in [-2\pi, \frac{5\pi}{2}] $

Solutions for $ n=3 $ ($ x = \frac{17\pi}{6} \approx 2.83\pi $) and $ n=-3 $ ($ x = -\frac{19\pi}{6} \approx -3.17\pi $) lie outside the specified interval.

Final Count

The distinct solutions for $ x $ in the interval $ [-2\pi, \frac{5\pi}{2}] $ are $ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6} $.

There are 5 solutions.