find all solutions of the equation in the interval 0, 2π).\ncos²x = 2 + 2sinx\nwrite your answer in radians…

find all solutions of the equation in the interval 0, 2π).\ncos²x = 2 + 2sinx\nwrite your answer in radians in terms of π.\nif there is more than one solution, separate them with commas.\nx =

find all solutions of the equation in the interval 0, 2π).\ncos²x = 2 + 2sinx\nwrite your answer in radians in terms of π.\nif there is more than one solution, separate them with commas.\nx =

Answer

Explanation:

Step1: Use trig identity $\cos^{2}x = 1-\sin^{2}x$

Substitute $\cos^{2}x$ in the given equation: $1 - \sin^{2}x=2 + 2\sin x$

Step2: Rearrange to form a quadratic equation

Move all terms to one - side: $\sin^{2}x+2\sin x + 1 = 0$ Let $t=\sin x$, then the equation becomes $t^{2}+2t + 1 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation: $(t + 1)^{2}=0$. So $t=\sin x=-1$.

Step4: Find $x$ in the given interval

Since $\sin x=-1$ and $x\in[0,2\pi)$, then $x=\frac{3\pi}{2}$.

Answer:

$x=\frac{3\pi}{2}$