Determine general solution \( 1+4 \sin ^{2} x-5 \sin x+\cos 2 x=0 \)

Solve the equation \( 1+4 \sin ^{2}(x)-5 \sin (x)+\cos (2x)=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(1+4\sin^{2}\left(x\right)-5\sin\left(x\right)+\cos\left(2x\right)=0\) - step1: Rewrite the expression: \(1+4\sin^{2}\left(x\right)-5\sin\left(x\right)+2\cos^{2}\left(x\right)-1=0\) - step2: Expand the expression: \(4\sin^{2}\left(x\right)-5\sin\left(x\right)+2\cos^{2}\left(x\right)=0\) - step3: Rewrite the expression: \(2\sin^{2}\left(x\right)-5\sin\left(x\right)+2=0\) - step4: Factor the expression: \(\left(\sin\left(x\right)-2\right)\left(2\sin\left(x\right)-1\right)=0\) - step5: Separate into possible cases: \(\begin{align}&\sin\left(x\right)-2=0\\&2\sin\left(x\right)-1=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x \notin \mathbb{R}\\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step7: 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 general solution to the equation \(1+4 \sin ^{2} x-5 \sin x+\cos 2x=0\) is given by: \[x = \left\{ \begin{array}{l} \frac{\pi}{6} + 2k\pi \\ \frac{5\pi}{6} + 2k\pi \end{array} \right., k \in \mathbb{Z}\] where \(k\) is an integer.