use an identity to solve the equation on the interval $0,2\\pi)$. $\\sin ^{2}x - 4\\cos x + 4 = 0$ what is…

use an identity to solve the equation on the interval $0,2\\pi)$. $\\sin ^{2}x - 4\\cos x + 4 = 0$ what is the solution in the interval $0,2\\pi)$? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. $x =$ (type your answer in radians. use integers or fractions for any numbers in the expression. type an exact answer, using $\\pi$ as needed. use a comma to separate answers as needed.) b. there is no solution.

use an identity to solve the equation on the interval $0,2\\pi)$. $\\sin ^{2}x - 4\\cos x + 4 = 0$ what is the solution in the interval $0,2\\pi)$? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. $x =$ (type your answer in radians. use integers or fractions for any numbers in the expression. type an exact answer, using $\\pi$ as needed. use a comma to separate answers as needed.) b. there is no solution.

Answer

Explanation:

Step1: Substitute the identity

We know that $\sin^{2}x = 1-\cos^{2}x$. Substitute this into the equation $\sin^{2}x - 4\cos x+4 = 0$. We get $1-\cos^{2}x-4\cos x + 4=0$.

Step2: Simplify the equation

Rearrange the terms: $-\cos^{2}x-4\cos x + 5=0$. Multiply through by $- 1$ to get $\cos^{2}x + 4\cos x-5=0$. Let $t=\cos x$, then the equation becomes $t^{2}+4t - 5=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $t^{2}+4t - 5=(t + 5)(t - 1)=0$. Set each factor equal to zero: $t+5=0$ gives $t=-5$ and $t - 1=0$ gives $t = 1$.

Step4: Back - substitute and find (x)

Since $t=\cos x$, when $\cos x=-5$, there is no solution because $-1\leqslant\cos x\leqslant1$. When $\cos x = 1$, and for $x\in[0,2\pi)$, we know that $x = 0$.

Answer:

A. (x = 0)