solve the equation on the interval 0,2π).\n\ncot x ( tan x + √3 ) = 0\n\nselect the correct choice below…

solve the equation on the interval 0,2π).\n\ncot x ( tan x + √3 ) = 0\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. x = □ (type an exact answer, using π as needed. use a comma to separate answers as needed. type your answer in radians. use integers or fractions for any numbers in the expression.)\n\nb. there is no solution.

solve the equation on the interval 0,2π).\n\ncot x ( tan x + √3 ) = 0\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. x = □ (type an exact answer, using π as needed. use a comma to separate answers as needed. type your answer in radians. use integers or fractions for any numbers in the expression.)\n\nb. there is no solution.

Answer

Explanation:

Step1: Use the zero - product property

If (ab = 0), then either (a=0) or (b = 0). For the equation (\cot x(\tan x+\sqrt{3})=0), we have two cases: Case 1: (\cot x=0), i.e., (\frac{\cos x}{\sin x}=0). This implies (\cos x = 0) and (\sin x\neq0). Case 2: (\tan x+\sqrt{3}=0), i.e., (\tan x=-\sqrt{3})

Step2: Solve (\cos x = 0) for (x\in[0,2\pi))

We know that (\cos x = 0) when (x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}). In the interval ([0,2\pi)), when (k = 0), (x=\frac{\pi}{2}); when (k = 1), (x=\frac{3\pi}{2})

Step3: Solve (\tan x=-\sqrt{3}) for (x\in[0,2\pi))

We know that (\tan x=\frac{\sin x}{\cos x}), and (\tan x =-\sqrt{3}) means (x=\arctan(-\sqrt{3})+k\pi). Since (\tan\frac{\pi}{3}=\sqrt{3}), then (\tan x=-\sqrt{3}) gives (x=\frac{2\pi}{3}+k\pi,k\in\mathbb{Z}). In the interval ([0,2\pi)), when (k = 0), (x=\frac{2\pi}{3}); when (k = 1), (x=\frac{5\pi}{3})

Answer:

(x=\frac{\pi}{2},\frac{2\pi}{3},\frac{3\pi}{2},\frac{5\pi}{3})