Question

Solve 4tanx/1-tan^2x=1 for 0 ≤ x ≤ 2π

Answer 1

x = 0.232, 1.803, 3.373, 4.944

Explanation:

Given trigonometric equation:

`(4 \tan x)/(1- \tan^2x)=1`

Multiply both sides of the equation by 1 - tan²x:

`4\tan x = 1 - \tan^2 x`

Add tan²x - 1 to both sides:

`\tan^2x+4\tan x-1=0`

Solve the quadratic equation by using the quadratic formula.

`\boxed{\begin{array}{c}\underline{\sf Quadratic\;Formula}\\\\x=(-b \pm √(b^2-4ac))/(2a)\quad\textsf{when}\;ax^2+bx+c=0\\\\\end{array}}`

In this case:

• a = 1
• b = 4
• c = -1

Substitute the values of a, b and c into the quadratic formula and solve for tan(x):

`\tan x=(-4\pm √(4^2-4(1)(-1)))/(2(1))`

`\tan x=(-4\pm √(16+4))/(2)`

`\tan x=(-4\pm √(20))/(2)`

`\tan x=(-4\pm √(2^2 \cdot 5))/(2)`

`\tan x=(-4\pm √(2^2) √(5))/(2)`

`\tan x=(-4\pm2√(5))/(2)`

`\tan x&=-2\pm√(5)`

So the two solutions for tan(x) are:

`\tan x=-2-√(5)`

`\tan x=-2+√(5)`

Solve for x by taking the arctan of both solutions, remembering that the tangent function is periodic with a period of π.

`x=\arctan\left(-2-√(5)\right)+\pi n\approx-1.339+\pi n`

`x=\arctan\left(-2+√(5)\right)+\pi n \approx0.232+\pi n`

Therefore, the solutions for x that fall within the given interval 0 ≤ x ≤ 2π are:

`x=\arctan\left(-2+√(5)\right) \approx0.232\; \sf (3\;d.p.)`

`x=\arctan\left(-2-√(5)\right)+\pi \approx 1.803 \; \sf \;(3\;d.p.)`

`x=\arctan\left(-2+√(5)\right)+\pi \approx 3.373\; \sf (3\;d.p.)`

`x=\arctan\left(-2-√(5)\right)+2\pi \approx 4.944\; \sf \;(3\;d.p.)`