Use trigonometric identities, algebraic methods, and inverse trigonometric functions, as necessary, to solve the following trigonometric equation on the interval [ \( 0,2 \pi \) ). Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution." \[ 2 \csc ^{2}(x)-22=6 \cot (x) \]

To solve the equation \( 2 \csc ^{2}(x)-22=6 \cot (x) \), we can start by utilizing the identity \( \csc^2(x) = 1 + \cot^2(x) \). 1. Substitute \( \csc^2(x) \): \[ 2(1 + \cot^2(x)) - 22 = 6 \cot(x) \] 2. Simplifying gives: \[ 2 + 2 \cot^2(x) - 22 = 6 \cot(x) \] \[ 2 \cot^2(x) - 6 \cot(x) - 20 = 0 \] 3. Now, we can divide the entire equation by 2 to simplify: \[ \cot^2(x) - 3 \cot(x) - 10 = 0 \] 4. Let's factor the quadratic equation: \[ (\cot(x) - 5)(\cot(x) + 2) = 0 \] 5. This gives us two cases: - \( \cot(x) - 5 = 0 \) → \( \cot(x) = 5 \) - \( \cot(x) + 2 = 0 \) → \( \cot(x) = -2 \) 6. Now, we find \( x \) for each case within the interval \( [0, 2\pi] \): - For \( \cot(x) = 5 \): \[ x = \cot^{-1}(5) \] This has one solution in the first quadrant and one in the third quadrant: \[ x_1 \approx \tan^{-1}(1/5) \approx 0.1974 \] \[ x_2 = \pi + 0.1974 \approx 3.3390 \] - For \( \cot(x) = -2 \): \[ x = \cot^{-1}(-2) \] This has one solution in the second quadrant and one in the fourth quadrant: \[ x_3 = \pi - \tan^{-1}(1/2) \approx 2.6780 \] \[ x_4 = 2\pi - \tan^{-1}(1/2) \approx 5.4633 \] 7. Thus, the solutions to the equation on the interval \([0, 2\pi]\) are: \[ x \approx 0.1974, \, 3.3390, \, 2.6780, \, 5.4633 \] Final rounded solutions (to four decimal places): \[ x \approx 0.1974, 2.6780, 3.3390, 5.4633 \]