which shows all the exact solutions of \\( \\sin ^{2}x + 3\\cos x - 1 = 2 \\)? give your answer in…

Explanation:

Step1: Use Pythagorean identity

We know that (\sin^{2}x = 1-\cos^{2}x). Substitute this into the equation (\sin^{2}x+3\cos x - 1=2): [ \begin{align*} 1-\cos^{2}x + 3\cos x-1&=2\ -\cos^{2}x+3\cos x - 2&=0\ \cos^{2}x-3\cos x + 2&=0 \end{align*} ]

Step2: Factor the quadratic equation

Let (t = \cos x). Then the equation (t^{2}-3t + 2=0) factors as ((t - 1)(t - 2)=0). So (t=\cos x=1) or (t=\cos x = 2). But (- 1\leqslant\cos x\leqslant1), so (\cos x=2) is rejected.

Step3: Solve (\cos x = 1)

The general solution of (\cos x=1) is (x = 2k\pi), where (k\in\mathbf{Z}) (using the property that (\cos\theta = 1) when (\theta = 2k\pi,k\in\mathbf{Z}))

Answer:

(2k\pi)