Question

Number of solutions of the equation \[ \sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0 \] in \( \theta \in \left[ -3\pi, 2\pi \right] \):

Hint:

For trigonometric equations involving double angles, use trigonometric identities to simplify the equation and then solve for the variable using the quadratic formula if applicable.

Answer 1

Correct Answer: 5

Simplifying the given equation involves trigonometric identities:
\[ \sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0 \] Use the identity \( \cos 2\theta = 2\cos^2\theta - 1 \):
\[ \sqrt{3}(2\cos^2\theta - 1) + 8\cos\theta + 3\sqrt{3} = 0 \] Simplify this:
\[ 2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0 \] Divide by 2:
\[ \sqrt{3}\cos^2\theta + 4\cos\theta + \sqrt{3} = 0 \] Let \( x = \cos\theta \). We get a quadratic:
\[ \sqrt{3}x^2 + 4x + \sqrt{3} = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = \sqrt{3}, b = 4, c = \sqrt{3} \):
\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2\sqrt{3}} = \frac{-4 \pm 2}{2\sqrt{3}} \] Compute roots:
\[ x_1 = \frac{-4 + 2}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \] \[ x_2 = \frac{-4 - 2}{2\sqrt{3}} = \frac{-6}{2\sqrt{3}} = -\frac{\sqrt{3}}{\sqrt{3}} = -1 \] These correspond to \( \cos\theta = -\frac{1}{\sqrt{3}} \) and \( \cos\theta = -1 \).
For \( \cos\theta = -\frac{1}{\sqrt{3}} \), \( \theta = 150^\circ, 210^\circ \) (or radians: \( \frac{5\pi}{6}, \frac{7\pi}{6} \)).
In \( \theta \in [-3\pi, 2\pi] \):
Periodicity of cosine \( 2\pi \) involves theta values:
\[ \theta = -3\pi + \frac{5\pi}{6}, -3\pi + \frac{7\pi}{6}, -\pi + \frac{5\pi}{6}, -\pi + \frac{7\pi}{6} \] For \( \cos\theta = -1 \), capture \( \theta = 180^\circ = \pi, -\pi \).
Valid theta values:
\[ \theta = -3\pi + \pi, -\pi, \pi, 3\pi \] Count total distinct theta values in interval:
\[ \text{Solutions: } 5 \]
Confirms within provided range 5,5.