6) Solve the trigonometric equations over the interval \( 0 \leq x \leq 2 \pi \) a) \( 2 \sin 2 x-\sin x=0 \) b) \( \cos 2 x+\sin ^{2} x-4 \cos x+3=0 \) c) \( 2 \csc x^{2}-8=0 \) ( \( 2=0 \)

Solve the equation \( 2 \sin 2x - \sin x = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2\sin\left(2x\right)-\sin\left(x\right)=0\) - step1: Rewrite the expression: \(2\times 2\sin\left(x\right)\cos\left(x\right)-\sin\left(x\right)=0\) - step2: Simplify: \(4\sin\left(x\right)\cos\left(x\right)-\sin\left(x\right)=0\) - step3: Factor the expression: \(\sin\left(x\right)\left(4\cos\left(x\right)-1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\sin\left(x\right)=0\\&4\cos\left(x\right)-1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=k\pi ,k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}-\arccos\left(\frac{1}{4}\right)+2k\pi \\\arccos\left(\frac{1}{4}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}-\arccos\left(\frac{1}{4}\right)+2k\pi \\k\pi \\\arccos\left(\frac{1}{4}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \cos 2x + \sin^{2}x - 4\cos x + 3 = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(2x\right)+\sin^{2}\left(x\right)-4\cos\left(x\right)+3=0\) - step1: Rewrite the expression: \(2\cos^{2}\left(x\right)-1+\sin^{2}\left(x\right)-4\cos\left(x\right)+3=0\) - step2: Expand the expression: \(2\cos^{2}\left(x\right)+2+\sin^{2}\left(x\right)-4\cos\left(x\right)=0\) - step3: Rewrite the expression: \(\cos^{2}\left(x\right)+3-4\cos\left(x\right)=0\) - step4: Factor the expression: \(\left(\cos\left(x\right)-3\right)\left(\cos\left(x\right)-1\right)=0\) - step5: Separate into possible cases: \(\begin{align}&\cos\left(x\right)-3=0\\&\cos\left(x\right)-1=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x \notin \mathbb{R}\\&x=2k\pi ,k \in \mathbb{Z}\end{align}\) - step7: Find the union: \(x=2k\pi ,k \in \mathbb{Z}\) Solve the equation \( 2 \csc x^{2} - 8 = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2\csc\left(x^{2}\right)-8=0\) - step1: Move the constant to the right side: \(2\csc\left(x^{2}\right)=0+8\) - step2: Remove 0: \(2\csc\left(x^{2}\right)=8\) - step3: Divide both sides: \(\frac{2\csc\left(x^{2}\right)}{2}=\frac{8}{2}\) - step4: Divide the numbers: \(\csc\left(x^{2}\right)=4\) - step5: Use the inverse trigonometric function: \(x^{2}=\operatorname{arccsc}\left(4\right)\) - step6: Calculate: \(\begin{align}&x^{2}=\operatorname{arccsc}\left(4\right)\\&x^{2}=-\operatorname{arccsc}\left(4\right)+\pi \end{align}\) - step7: Add the period: \(\begin{align}&x^{2}=\operatorname{arccsc}\left(4\right)+2k\pi ,k \in \mathbb{Z}\\&x^{2}=-\operatorname{arccsc}\left(4\right)+\pi +2k\pi ,k \in \mathbb{Z}\end{align}\) - step8: Calculate: \(\begin{align}&x=\left\{ \begin{array}{l}\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\\-\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\end{array}\right.,k \in \mathbb{Z}\\&x^{2}=-\operatorname{arccsc}\left(4\right)+\pi +2k\pi ,k \in \mathbb{Z}\end{align}\) - step9: Calculate: \(\begin{align}&x=\left\{ \begin{array}{l}\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\\-\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\end{array}\right.,k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}\sqrt{-\operatorname{arccsc}\left(4\right)+\pi +2k\pi }\\-\sqrt{-\operatorname{arccsc}\left(4\right)+\pi +2k\pi }\end{array}\right.,k \in \mathbb{Z}\end{align}\) - step10: Find the union: \(x=\left\{ \begin{array}{l}\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\\-\sqrt{\operatorname{arccsc}\left(4\right)+2k\pi }\\\sqrt{-\operatorname{arccsc}\left(4\right)+\pi +2k\pi }\\-\sqrt{-\operatorname{arccsc}\left(4\right)+\pi +2k\pi }\end{array}\right.,k \in \mathbb{Z}\) The solutions to the trigonometric equations over the interval \(0 \leq x \leq 2\pi\) are as follows: a) \(2 \sin 2x - \sin x = 0\) The solutions are: \[ x = -\arccos\left(\frac{1}{4}\right) + 2k\pi \] \[ x = k\pi \] \[ x = \arccos\left(\frac{1}{4}\right) + 2k\pi \] where \(k\) is an integer. b) \(\cos 2x + \sin^{2}x - 4\cos x + 3 = 0\) The solution is: \[ x = 2k\pi \] where \(k\) is an integer. c) \(2 \csc x^{2} - 8 = 0\) The solutions are: \[ x = \sqrt{\arccsc(4) + 2k\pi} \] \[ x = -\sqrt{\arccsc(4) + 2k\pi} \] \[ x = \sqrt{-\arccsc(4) + \pi + 2k\pi} \] \[ x = -\sqrt{-\arccsc(4) + \pi + 2k\pi} \] where \(k\) is an integer.