find all solutions of the equation in the interval 0, 2π). (2 sinx - √3)(cosx - 4)=0 write your answer in…

Explanation:

Step1: Set each factor equal to zero

If ((2\sin x-\sqrt{3})(\cos x - 4)=0), then either (2\sin x-\sqrt{3}=0) or (\cos x - 4=0).

Step2: Analyze (\cos x - 4=0)

Since (- 1\leqslant\cos x\leqslant1), the equation (\cos x-4 = 0) (i.e., (\cos x=4)) has no solutions.

Step3: Solve (2\sin x-\sqrt{3}=0)

First, rewrite (2\sin x-\sqrt{3}=0) as (\sin x=\frac{\sqrt{3}}{2}). We know that (\sin x=\frac{\sqrt{3}}{2}) when (x = \frac{\pi}{3}+2k\pi) or (x=\frac{2\pi}{3}+2k\pi), (k\in\mathbb{Z}).

Step4: Find solutions in ([0,2\pi))

For (k = 0), when (x=\frac{\pi}{3}) and (x=\frac{2\pi}{3}), they are in the interval ([0,2\pi)).

Answer:

(x=\frac{\pi}{3},\frac{2\pi}{3})