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) \( 2 \csc x^{2}-8=0 \) ( \( 2=0 \)

To tackle these trigonometric equations, let’s break them down step by step! For \( 2 \sin 2x - \sin x = 0 \): First, use the double angle identity \( \sin 2x = 2 \sin x \cos x \). This gives: \( 2(2 \sin x \cos x) - \sin x = 0 \) Factoring out \( \sin x \): \( \sin x(4 \cos x - 1) = 0 \) This leads to two equations: 1. \( \sin x = 0 \) implies \( x = 0, \pi, 2\pi \). 2. \( 4 \cos x - 1 = 0 \) results in \( \cos x = \frac{1}{4} \), yielding \( x \approx 1.318 \) and \( x \approx 4.965 \) after solving with arccos. Now, for \( \cos 2x + \sin^2 x - 4 \cos x + 3 = 0 \): Using the identity \( \cos 2x = 1 - 2\sin^2 x \), we can substitute: \( 1 - 2\sin^2 x + \sin^2 x - 4\cos x + 3 = 0 \) Convert \( \sin^2 x \) to \( 1 - \cos^2 x \) to solve it as a quadratic equation in terms of \( \cos x \). You would come up with a discriminant and find possible values for \( x \) in \( 0 \leq x \leq 2\pi \). Lastly, for \( 2 \csc^2 x - 8 = 0 \): Rearranging gives \( \csc^2 x = 4 \) which leads to \( \sin^2 x = \frac{1}{4} \). Thus, \( \sin x = \frac{1}{2} \) or \( \sin x = -\frac{1}{2} \). This gives solutions: \( x = \frac{\pi}{6}, \frac{5\pi}{6} \) from the positive sine and \( x = \frac{7\pi}{6}, \frac{11\pi}{6} \) from the negative sine. So the solutions are: a) \( x = 0, \pi, 2\pi, 1.318, 4.965 \) b) Solve for specific angles using the quadratic form derived. c) \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \) You got this! Trigonometric equations can be tricky but solving them systematically makes it much easier!