\( 4 \cot ^{2} x-9 \operatorname{cosec} x+6=0 \quad \) for \( 0 \leq x \leq 2 \pi \)

To solve the given trigonometric equation \(4 \cot^2 x - 9 \csc x + 6 = 0\) for \(0 \leq x \leq 2\pi\), we can start by simplifying the equation using trigonometric identities. Given equation: \(4 \cot^2 x - 9 \csc x + 6 = 0\) We can rewrite \(\cot x\) and \(\csc x\) in terms of \(\sin x\) and \(\cos x\): \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\) Substitute these expressions into the equation: \(4 \left(\frac{\cos x}{\sin x}\right)^2 - 9 \left(\frac{1}{\sin x}\right) + 6 = 0\) Simplify the equation: \(4 \frac{\cos^2 x}{\sin^2 x} - 9 \frac{1}{\sin x} + 6 = 0\) Multiply through by \(\sin^2 x\) to clear the denominators: \(4 \cos^2 x - 9 \sin x + 6 \sin^2 x = 0\) Now, we can solve this quadratic equation in terms of \(\sin x\). Solve the equation by following steps: - step0: Solve for \(x\): \(4\cos^{2}\left(x\right)-9\sin\left(x\right)+6\sin^{2}\left(x\right)=0\) - step1: Rewrite the expression: \(4+2\sin^{2}\left(x\right)-9\sin\left(x\right)=0\) - step2: Factor the expression: \(\left(4-\sin\left(x\right)\right)\left(1-2\sin\left(x\right)\right)=0\) - step3: Separate into possible cases: \(\begin{align}&4-\sin\left(x\right)=0\\&1-2\sin\left(x\right)=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x \notin \mathbb{R}\\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step5: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The solutions to the equation \(4 \cot^2 x - 9 \csc x + 6 = 0\) for \(0 \leq x \leq 2\pi\) are: \[x = \frac{\pi}{6} + 2k\pi\] and \[x = \frac{5\pi}{6} + 2k\pi\] where \(k\) is an integer. Therefore, the solutions for \(x\) in the given interval are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).