for \\( \\sin 2 x + \\cos x = 0 \\), use a double - angle or half - angle formula to simplify the equation…

Explanation:

Step1: Apply double - angle formula

Use the double - angle formula (\sin2x = 2\sin x\cos x). The equation (\sin2x+\cos x = 0) becomes (2\sin x\cos x+\cos x=0). Factor out (\cos x): (\cos x(2\sin x + 1)=0).

Step2: Solve (\cos x=0)

If (\cos x = 0), then (x=\frac{\pi}{2}+k\pi), (k\in\mathbb{Z}). For (x\in[0,2\pi)), when (k = 0), (x=\frac{\pi}{2}); when (k = 1), (x=\frac{3\pi}{2}).

Step3: Solve (2\sin x+1 = 0)

If (2\sin x+1=0), then (\sin x=-\frac{1}{2}). The solutions of (\sin x =-\frac{1}{2}) for (x\in[0,2\pi)) are (x=\frac{7\pi}{6}) and (x=\frac{11\pi}{6}) (since (\sin x=y) and (y =-\frac{1}{2}), and using the unit - circle definition of sine where (\sin x) is the (y) - coordinate of a point ((\cos x,\sin x)) on the unit circle (x^{2}+y^{2}=1)).

Answer:

(x_1=\frac{\pi}{2}), (x_2=\frac{7\pi}{6}), (x_3=\frac{3\pi}{2}), (x_4=\frac{11\pi}{6})