Find all solutions of the equation in the interval [0, 2π]. √3 csc(2x) + 2 = 0. Write your answer in radians in terms of π.

Steps to Solve

1. Isolate the csc(2x) term

Start with the original equation: $$ \sqrt{3} \csc(2x) + 2 = 0 $$

Subtract 2 from both sides: $$ \sqrt{3} \csc(2x) = -2 $$

Now, divide by $\sqrt{3}$: $$ \csc(2x) = -\frac{2}{\sqrt{3}} $$

2. Convert csc to sin

Recall that $\csc(2x)$ is the reciprocal of $\sin(2x)$: $$ \frac{1}{\sin(2x)} = -\frac{2}{\sqrt{3}} $$

Therefore, $$ \sin(2x) = -\frac{\sqrt{3}}{2} $$

3. Find angles where sine equals -√3/2

The sine function equals $-\frac{\sqrt{3}}{2}$ at specific angles in the unit circle: $$ 2x = \frac{4\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{3} + 2k\pi $$

for integers $k$.

4. Solve for x by dividing all terms by 2

To find $x$, divide the equations by 2: $$ x = \frac{2\pi}{3} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi $$

5. Determine values within [0, 2π]

Substituting $k = 0$ and $k = 1$ to find valid $x$ values:

• For $x = \frac{2\pi}{3}$:

  • $k = 0 \implies \frac{2\pi}{3}$
  • $k = 1 \implies \frac{2\pi}{3} + \pi = \frac{5\pi}{3}$

• For $x = \frac{5\pi}{6}$:

  • $k = 0 \implies \frac{5\pi}{6}$
  • $k = 1 \implies \frac{5\pi}{6} + \pi = \frac{11\pi}{6}$

All values in the interval [0, 2π] are: $$ \frac{2\pi}{3}, \frac{5\pi}{6}, \frac{5\pi}{3}, \frac{11\pi}{6} $$