Solve the trigonometric equation for all values $$ 0 \leq x

Question

Solve the trigonometric equation for all values $$ 0 \leq x<2 \pi $$ $$ 2 \cos x=-\sqrt{3} $$

UpStudy AI · Accepted Answer

1. Start with the equation:
   $$
   2\cos x = -\sqrt{3}
   $$

2. Divide both sides by 2 to isolate $$\cos x$$:
   $$
   \cos x = -\frac{\sqrt{3}}{2}
   $$

3. Identify the angles in the interval $$ 0 \leq x < 2\pi $$ where $$\cos x = -\frac{\sqrt{3}}{2}$$. Recall that:
   - $$\cos x = \frac{\sqrt{3}}{2}$$ at $$x = \frac{\pi}{6}$$ and $$x = \frac{11\pi}{6}$$,
   - And the cosine function is negative in the second and third quadrants.

4. The corresponding angles in the second and third quadrants are:
   - Second quadrant: $$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$
   - Third quadrant: $$x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$

5. Therefore, the solutions for $$ x $$ are:
   $$
   x = \frac{5\pi}{6} \quad \text{and} \quad x = \frac{7\pi}{6}
   $$
The solutions are $$ x = \frac{5\pi}{6} $$ and $$ x = \frac{7\pi}{6} $$.

Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ 2 \cos x=-\sqrt{3} \]

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