Question

Question #5a4e4

Answer 1

`x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6`

Explanation:

We have the equation:

`6sec^2x+3tan^2x-9=0`

The Pythagorean identity can give us the following relation between `sec^2x` and `tan^2x`:

`tan^2x+1=sec^2x`

Thus, in the first equation, we can replace `sec^2x` with `(tan^2x+1)`, giving the new equation:

`6(tan^2x+1)+3tan^2x-9=0`

Divide both sides of the equation by `3`:

`2(tan^2x+1)+tan^2x-3=0`

Distribute the `2`.

`2tan^2x+2+tan^2x-3=0`

Combine like terms.

`3tan^2x-1=0`

Add `1` to both sides of the equation.

`3tan^2x=1`

Divide both sides of the equation by `3`.

`tan^2x=1/3`

Take the square root of both sides of the equation. Recall that the positive and negative versions are valid.

`tanx=+-sqrt(1/3)`

Note that `sqrt(1/3)=1/sqrt3=sqrt3/3`.

`tanx=+-sqrt3/3`

This is a commonly known value of tangent, that is, you should know that `tan(pi/6)=sqrt3/3`.

Since the positive and negative versions of this are allowed and the domain is `0<x<2pi`, the four solutions are the four angles in each quadrant with a reference angle of `pi/6`, which are:

`x=overbrace(0+pi/6)^"QI",overbrace(pi-pi/6)^"QII",overbrace(pi+pi/6)^"QIII",overbrace(2pi-pi/6)^"QIV"`

Which, when simplified, give

`x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6`