Question

Consider the equation 3tan^2(x)−1=0. the solutions of this equation over the interval {0,pi} are?

Answer 1

`x = (\pi)/(6)`.

Explanation:

We have the following equation:

`3tan^2x - 1 =0`

`3tan^2x=1`

`tan^2x=(1)/(3)`

We use the property
`tanx = (sinx)/(cosx)` in our equation as follows:

`(sin^2x)/(cos^2x)=(1)/(3)`

and then, we use the identity
`cos^2x = 1 - sin^2x`:

`sin^2x=(1)/(3)cos^2x`

`sin^2x=(1)/(3)(1- sin^2x)`

`sin^2x=(1)/(3) - (1)/(3)sin^2x`

`sin^2x+(1)/(3)sin^2x=(1)/(3)`

`(4)/(3)sin^2x = (1)/(3)`

`sin^2x = (1)/(4)`

`sinx = \sqrt{(1)/(4)}`

`sinx =\pm(1)/(2)`

As we are over the interval (0,pi) we use the positive value:

`sinx =(1)/(2)`

`x =sin^(-1)((1)/(2))`

`x = (\pi)/(6)`.

Answer 2

If we solve for tan(x) algebraically, we are then able to refer to the unit circle to find the solutions. Solving for tan(x) algebraically will give us tan(x) = (square root of 3)/3. On the unit circle, there are two possibilities with those coordinates: Pi/6 and 7pi/6