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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 x1 be an angle in Quadrant II. The terminal side of the angle x1 lies in Quadrant II. P is a point on the terminal side of the angle x1. P has coordinate (x, y). Then tan x1 = y/x. In Quadrant II, x is a negative number and y is a positive number, so, tan x1 is a negative number. So, there is an angle x1 that will satisfy tan x = -3.

Let x2 be an angle in Quadrant IV. The terminal side of the angle x2 lies in Quadrant IV. P is a point on the terminal side of the angle x2. P has coordinate (x, y). Then tan x2 = y/x. In Quadrant IV, x is a positive number and y is a negative number, so, tan x2 is a negative number. So, there is an angle x2 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 = 71.57o

The angle in Quadrant II is: x1 = 180o - 71.57o = 108.43o

The angle in Quadrant IV is: x2 = 360o - 71.57o = 288.43o

Therefore, the set of the angles that satisfy tan x = - 3 and the angle x range from 0 to 2pi is: {108.43o, 288.43o} 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 x1 and x2. Angle x1 lies in Quadrant II and angle x2 lies in Quadrant IV. x1 and x2 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 {108.43o, 288.43o}. For more details, please watch the video.