Question

4tantheta/12-tan^2=1

Answer 1

`4tan \theta = 12 - tan^(2) \theta`

`tan^(2) \theta + 4tan \theta - 12 = 0`

Let
`x = tan \theta`

`x^(2) + 4x - 12 = 0`

`(x + 6)(x - 2) = 0`

`x = -6, x = 2`


`tan \theta = -6, tan \theta = 2`

Since they both work, then we can write the solutions in general form:

`\theta = n \pi + tan^(-1)(2)`,
`n \in Z`

`\theta = n \pi - tan^(-1)(6)`,
`n \in Z`

Now, these solutions will always satisfy the equation and substituting a whole number in place of n will produce a solution that satisfies the given equation.