use an identity to solve the equation on the interval 0,2π).\n\\( \\sin ^ { 2 } x - 5 \\cos x + 5 = 0…

use an identity to solve the equation on the interval 0,2π).\n\\( \\sin ^ { 2 } x - 5 \\cos x + 5 = 0 \\)\nwhat is the solution in the interval 0,2π)?\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. \\( x = \\) (type your answer in radians. use integers or fractions for any numbers in the expression. type an exact answer\nb. there is no solution.

use an identity to solve the equation on the interval 0,2π).\n\\( \\sin ^ { 2 } x - 5 \\cos x + 5 = 0 \\)\nwhat is the solution in the interval 0,2π)?\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. \\( x = \\) (type your answer in radians. use integers or fractions for any numbers in the expression. type an exact answer\nb. there is no solution.

Answer

Explanation:

Step1: Use the Pythagorean identity

We know that (\sin^{2}x = 1-\cos^{2}x). Substitute this into the given equation (\sin^{2}x-5\cos x + 5=0). We get (1-\cos^{2}x-5\cos x + 5=0).

Step2: Simplify the equation

Rearrange the terms: (-\cos^{2}x-5\cos x+6 = 0). Multiply through by (- 1) to get (\cos^{2}x + 5\cos x-6=0). Let (t=\cos x), then the equation becomes (t^{2}+5t - 6=0).

Step3: Solve the quadratic equation

Factor the quadratic equation (t^{2}+5t - 6=(t + 6)(t - 1)=0). Set each factor equal to zero: (t+6 = 0) gives (t=-6), and (t - 1=0) gives (t = 1).

Step4: Substitute back (t=\cos x)

Since (-1\leqslant\cos x\leqslant1), (\cos x=-6) is rejected. For (\cos x=1), and (x\in[0,2\pi)), we know that (x = 0).

Answer:

A. (x = 0)