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# How to find the angle set when given value of tan x?

Quastion:

Given: tan x = - 3, in which x is from 0 to 2pi. Find the set of the angle x.

Solution:

Because tan x = -3 which is a negative number, so, the angle x lies in Quadrant II or Quadrant IV.

Let x_{1} be an angle in Quadrant II. The terminal side of the angle x_{1} lies in Quadrant II. P is a point on the terminal side of the angle x_{1}. P has coordinate (x, y).
Then tan x_{1} = y/x. In Quadrant II, x is a negative number and y is a positive number, so, tan x_{1} is a negative number. So, there is an angle x_{1} that will satisfy tan x = -3.

Let x_{2} be an angle in Quadrant IV. The terminal side of the angle x_{2} lies in Quadrant IV. P is a point on the terminal side of the angle x_{2}. P has coordinate (x, y).
Then tan x_{2} = y/x. In Quadrant IV, x is a positive number and y is a negative number, so, tan x_{2} is a negative number. So, there is an angle x_{2} that will satisfy tan x = -3.

Now we will find an acute angle that corresponding to the absolute value of given trigonometry function. That is, let tan x = |- 3| = 3. Then x = arc tan 3 = 71.57^{o}

The angle in Quadrant II is: x_{1} = 180^{o} - 71.57^{o} = 108.43^{o}

The angle in Quadrant IV is: x_{2} = 360^{o} - 71.57^{o} = 288.43^{o}

Therefore, the set of the angles that satisfy tan x = - 3 and the angle x range from 0 to 2pi is: {108.43^{o}, 288.43^{o}}

Look the figure above, the blue curve is the graph of y = tan x. the orange line is y = -3. The intersection point of the blue curve and orange line is x_{1} and x_{2}.
Angle x_{1} lies in Quadrant II and angle x_{2} lies in Quadrant IV. x_{1} and x_{2} are the set of the angles that satisfy tan x = -3 and the angle x range from 0 to 2pi.
The set of the angles is {108.43^{o}, 288.43^{o}}. For more details, please watch the video.

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