solve the equation on the interval $0,2\\pi)$. \n$5\\cos x - \\sqrt{3} = 3\\cos x$ \n$x = \\frac{\\pi}{?}…

Explanation:

Step1: Simplify the equation

Subtract (3\cos x) from both sides: (5\cos x - 3\cos x-\sqrt{3}=3\cos x - 3\cos x) (2\cos x-\sqrt{3}=0) Then add (\sqrt{3}) to both sides: (2\cos x=\sqrt{3}), so (\cos x=\frac{\sqrt{3}}{2})

Step2: Find the values of (x) in ([0,2\pi))

We know that (\cos x = \frac{\sqrt{3}}{2}) when (x=\frac{\pi}{6}) (since (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2})) and (x = 2\pi-\frac{\pi}{6}=\frac{11\pi}{6}) (because the cosine function is positive in the first and fourth quadrants and (\cos(2\pi - \alpha)=\cos\alpha))

Answer:

(x=\frac{\pi}{6},\frac{11\pi}{6})