What are the solutions of 6 tanx sinx6 tanx=0 in the interval [0,2pi]?
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To solve the equation (6 \tan x \sin x  6 \tan x = 0) in the interval ([0, 2\pi]), we can first factor out (6 \tan x) to get:
[6 \tan x (\sin x  1) = 0.]
This equation is satisfied if either (6 \tan x = 0) or (\sin x  1 = 0).

(6 \tan x = 0): This occurs when (\tan x = 0). In the interval ([0, 2\pi]), the solutions for (\tan x = 0) are (x = 0, \pi, 2\pi).

(\sin x  1 = 0): This occurs when (\sin x = 1). In the interval ([0, 2\pi]), the solution for (\sin x = 1) is (x = \frac{\pi}{2}).
Therefore, the solutions of the equation (6 \tan x \sin x  6 \tan x = 0) in the interval ([0, 2\pi]) are (x = 0, \frac{\pi}{2}, \pi, 2\pi).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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