-4\\sin^{2}x + 7\\cos x + 2 = 0

Explanation:

Step1: Use trigonometric identity

Since (\sin^{2}x = 1-\cos^{2}x), the equation (- 4\sin^{2}x+7\cos x + 2 = 0) can be rewritten as (-4(1 - \cos^{2}x)+7\cos x + 2 = 0). Expand the left - hand side: (-4 + 4\cos^{2}x+7\cos x + 2 = 0). Simplify to get a quadratic equation in terms of (\cos x): (4\cos^{2}x+7\cos x - 2 = 0).

Step2: Let (t=\cos x)

Let (t = \cos x), then the equation becomes (4t^{2}+7t - 2=0). Use the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (at^{2}+bt + c = 0). Here (a = 4), (b = 7), (c=-2). First, calculate the discriminant (\Delta=b^{2}-4ac=(7)^{2}-4\times4\times(-2)=49 + 32=81). Then (t=\frac{-7\pm\sqrt{81}}{2\times4}=\frac{-7\pm9}{8}). We have two solutions for (t): (t_{1}=\frac{-7 + 9}{8}=\frac{2}{8}=\frac{1}{4}) and (t_{2}=\frac{-7-9}{8}=\frac{-16}{8}=-2).

Step3: Solve for (x)

Since (t = \cos x) and (-1\leqslant\cos x\leqslant1), we discard (t=-2) (because (\cos x=-2) has no solution). When (\cos x=\frac{1}{4}), then (x = 2k\pi\pm\arccos(\frac{1}{4})), where (k\in\mathbb{Z}).

Answer:

(x = 2k\pi\pm\arccos(\frac{1}{4}),k\in\mathbb{Z})