Solving the Equation tan2 x - 5 tan x - 6 0 for x where 0 ≤ x ≤ 2π

Introduction

In mathematics, solving trigonometric equations is an important aspect of algebra and calculus. A common method involves the use of substitution and factoring techniques. This article will guide you through the process of solving the equation tan2x - 5tan x - 6 0 for x within the interval 0 ≤ x ≤ 2π.

Method of Solution via Substitution

To solve the equation tan2x - 5tan x - 6 0, we start by making a substitution:

Let u tan x

Substituting u into the equation, we get:

u2 - 5u - 6 0

This is a quadratic equation, which can be factored as:

(u - 6)(u 1) 0

Setting each factor to zero gives us two potential solutions for u:

u - 6 0 ? u 6 u 1 0 ? u -1

Substituting back in for u tan x, we have:

tan x -1 tan x 6

To find the corresponding values of x, we need to use the inverse tangent function, tan-1, and consider the given interval 0 ≤ x ≤ 2π.

Calculating the Solutions

For tan x -1, the solutions within the interval 0 ≤ x ≤ 2π are:

x tan-1(-1) kπ -π/4 kπ

Considering the values of k (0, 1) within the interval, we get:

x -π/4 0π -π/4 π 3π/4 x -π/4 1π 3π/4 π 7π/4

For tan x 6, the solutions are:

x tan-1(6) kπ

Considering the values of k (0, 1) within the interval, we get:

x tan-1(6) 0π tan-1(6) x tan-1(6) 1π tan-1(6) π

Conclusion

The complete solution to the equation tan2x - 5tan x - 6 0 within the interval 0 ≤ x ≤ 2π is:

x 3π/4, 7π/4, tan-1(6), tan-1(6) π

Using the substitution and inverse tangent functions, we have successfully solved the given equation.