Solve the Trigonometric Equation

The given equation is:

$\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0$

The interval is $\theta \in [-3\pi, 2\pi]$.

• Step 1: Use the double angle identity

  Substitute $\cos 2\theta = 2\cos^2 \theta - 1$ into the equation:

  $\sqrt{3}(2\cos^2 \theta - 1) + 8\cos \theta + 3\sqrt{3} = 0$

• Step 2: Simplify to a quadratic form

  Expand and rearrange the terms:

  $2\sqrt{3}\cos^2 \theta - \sqrt{3} + 8\cos \theta + 3\sqrt{3} = 0$

  $2\sqrt{3}\cos^2 \theta + 8\cos \theta + 2\sqrt{3} = 0$

  Divide by 2:

  $\sqrt{3}\cos^2 \theta + 4\cos \theta + \sqrt{3} = 0$

• Step 3: Solve the quadratic equation for $\cos \theta$

  Let $x = \cos \theta$. The equation becomes $\sqrt{3}x^2 + 4x + \sqrt{3} = 0$.

  Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

  $x = \frac{-4 \pm \sqrt{4^2 - 4(\sqrt{3})(\sqrt{3})}}{2\sqrt{3}}$

  $x = \frac{-4 \pm \sqrt{16 - 12}}{2\sqrt{3}} = \frac{-4 \pm \sqrt{4}}{2\sqrt{3}} = \frac{-4 \pm 2}{2\sqrt{3}}$

  The possible values for x are:

  $x_1 = \frac{-4 + 2}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}}$

  $x_2 = \frac{-4 - 2}{2\sqrt{3}} = \frac{-6}{2\sqrt{3}} = -\sqrt{3}$

• Step 4: Determine valid solutions for $\cos \theta$

  Since $-1 \le \cos \theta \le 1$:

  • $\cos \theta = -\frac{1}{\sqrt{3}}$ is valid (approximately -0.577).
  • $\cos \theta = -\sqrt{3}$ is invalid (approximately -1.732), as it is less than -1.
• Step 5: Find solutions for $\theta$ in the given interval

  We need to find solutions for $\cos \theta = -\frac{1}{\sqrt{3}}$ in $\theta \in [-3\pi, 2\pi]$.

  Let $\alpha = \arccos\left(-\frac{1}{\sqrt{3}}\right)$. Note that $\alpha$ is in the second quadrant, i.e., $\alpha \in (\frac{\pi}{2}, \pi)$.

  The general solutions are of the form $\theta = 2k\pi \pm \alpha$, where $k$ is an integer.

  By analyzing the interval $[-3\pi, 2\pi]$ (which has a length of $5\pi$), we find the solutions.

  Checking integer values for $k$:

  • For $k = -1$: $\theta = -2\pi \pm \alpha$. Both $-2\pi + \alpha$ and $-2\pi - \alpha$ fall within the interval.
  • For $k = 0$: $\theta = \pm \alpha$. Both $\alpha$ and $-\alpha$ fall within the interval.
  • For $k = 1$: $\theta = 2\pi \pm \alpha$. Only $2\pi - \alpha$ falls within the interval (as $2\pi + \alpha$ is greater than $2\pi$).

  Counting these valid solutions within the specified interval $[-3\pi, 2\pi]$, we find there are 4 solutions.