find all solutions of the equation in the interval $0,2\\pi)$. $\\sin ^{2}x = 2 + 2\\cos x$ write your…

find all solutions of the equation in the interval $0,2\\pi)$. $\\sin ^{2}x = 2 + 2\\cos x$ write your answer in radians in terms of $\\pi$. if there is more than one solution, separate them with commas. $x = \\square$

find all solutions of the equation in the interval $0,2\\pi)$. $\\sin ^{2}x = 2 + 2\\cos x$ write your answer in radians in terms of $\\pi$. if there is more than one solution, separate them with commas. $x = \\square$

Answer

Explanation:

Step1: Use Pythagorean identity

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

Step2: Factor the quadratic equation

The quadratic equation (\cos^{2}x+2\cos x + 1 = 0) can be factored as ((\cos x + 1)^{2}=0).

Step3: Solve for (\cos x)

Take the square root of both sides: (\cos x+1 = 0), so (\cos x=-1).

Step4: Find (x) in the interval ([0,2\pi))

We know that (\cos x=-1) when (x = \pi) in the interval ([0,2\pi)).

Answer:

(\pi)