find the general solution (in radians) of the trigonometric equation below for all real values of x…

find the general solution (in radians) of the trigonometric equation below for all real values of x, expressing your answer in terms of k as determined by the dropdown below. 2 cos x - √3 = 0 answer attempt 1 out of 2 use the button below to add a second expression if necessary. x =

find the general solution (in radians) of the trigonometric equation below for all real values of x, expressing your answer in terms of k as determined by the dropdown below. 2 cos x - √3 = 0 answer attempt 1 out of 2 use the button below to add a second expression if necessary. x =

Answer

Explanation:

Step1: Isolate cosine function

First, solve the equation (2\cos x-\sqrt{3} = 0) for (\cos x). Add (\sqrt{3}) to both sides and then divide by 2. (\cos x=\frac{\sqrt{3}}{2})

Step2: Recall cosine - angle relationship

We know that (\cos x=\frac{\sqrt{3}}{2}) has solutions based on the unit - circle. The principal values of (x) for which (\cos x=\frac{\sqrt{3}}{2}) are (x = \frac{\pi}{6}) and (x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}) in the interval ([0, 2\pi]). The general solution of the cosine equation (\cos x = a), where (|a|\leq1), is given by (x = 2k\pi\pm\alpha), where (\alpha) is the principal value of (x) such that (\cos\alpha=a). Here, (\alpha=\frac{\pi}{6}), so the general solution is (x = 2k\pi\pm\frac{\pi}{6},k\in\mathbb{Z}).

Answer:

(x = 2k\pi+\frac{\pi}{6},x = 2k\pi-\frac{\pi}{6},k\in\mathbb{Z})