tan^2 x − 7 tan x − 8 = 0
Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π).
(tanx-8)(tanx+1) = 0
tanx = 8 or -1
arctan(8) = 1.446 or 4.588
arctan(-1) = 3π/4 or 7π/4
To find the solutions of the equation tan^2(x) - 7tan(x) - 8 = 0 in the interval [0, 2π), we can use inverse trigonometric functions.
Step 1: Convert the equation to a quadratic equation:
Let's rewrite the equation in terms of tan(x):
tan^2(x) - 7tan(x) - 8 = 0
Step 2: Factor the quadratic equation:
(tan(x) - 8)(tan(x) + 1) = 0
Step 3: Set each factor equal to zero and solve for x:
tan(x) - 8 = 0 or tan(x) + 1 = 0
For tan(x) - 8 = 0:
tan(x) = 8
Since we want to find solutions in the interval [0, 2π), we need to find the reference angle that corresponds to tan(8). Using the inverse tangent (atan) function, we can find the reference angle:
atan(8) ≈ 1.446
But we need to determine in which quadrants the angle has the given tangent value (8). Since tangent is positive in the first and third quadrants, we find another angle that also has a tangent of 8 in the third quadrant.
۞ Angle in Q1: atan(8) ≈ 1.446
۞ Angle in Q3: atan(8) + π ≈ 1.446 + π ≈ 4.587
So, we have two angles that have a tangent of 8: x ≈ 1.446 and x ≈ 4.587.
For tan(x) + 1 = 0:
tan(x) = -1
Again, we use the inverse tangent function to find the reference angle:
atan(-1) ≈ -0.785
But we need to determine in which quadrants the angle has the given tangent value (-1). Since tangent is negative in the second and fourth quadrants, we find another angle that also has a tangent of -1 in the fourth quadrant.
۞ Angle in Q2: atan(-1) + π ≈ -0.785 + π ≈ 2.356
۞ Angle in Q4: atan(-1) + π + π ≈ -0.785 + π + π ≈ 5.498
So, we have two more angles that have a tangent of -1: x ≈ 2.356 and x ≈ 5.498.
Therefore, the solutions to the equation tan^2(x) - 7tan(x) - 8 = 0 in the interval [0, 2π) are:
x ≈ 1.446, x ≈ 2.356, x ≈ 4.587, and x ≈ 5.498.