Question

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

√3 tan(x- π/8)-1=0

Answer 1

`(7\pi)/(24)` and
`(31\pi)/(24)`

Explanation:

`√(3) \tan(x-(\pi)/(8))-1=0`

Let's first isolate the trig function.

Add 1 one on both sides:

`√(3) \tan(x-(\pi)/(8))=1`

Divide both sides by
`√(3)`:

`\tan(x-(\pi)/(8))=(1)/(√(3))`

Now recall
`\tan(u)=(\sin(u))/(\cos(u))`.

`(1)/(√(3))=((1)/(2))/((√(3))/(2))`

or

`(1)/(√(3))=(-(1)/(2))/(-(√(3))/(2))`

The first ratio I have can be found using
`(\pi)/(6)` in the first rotation of the unit circle.

The second ratio I have can be found using
`(7\pi)/(6)` you can see this is on the same line as the
`(\pi)/(6)` so you could write
`(7\pi)/(6)` as
`(\pi)/(6)+\pi`.

So this means the following:

`\tan(x-(\pi)/(8))=(1)/(√(3))`

is true when
`x-(\pi)/(8)=(\pi)/(6)+n \pi`

where
`n` is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

`x-(\pi)/(8)=(\pi)/(6)+n \pi`

Add
`(\pi)/(8)` on both sides:

`x=(\pi)/(6)+(\pi)/(8)+n \pi`

Find common denominator between the first two terms on the right.

That is 24.

`x=(4\pi)/(24)+(3\pi)/(24)+n \pi`

`x=(7\pi)/(24)+n \pi` (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval
`[0,2\pi)`.

So if
`√(3) \tan(x-(\pi)/(8))-1=0` and we let
`u=x-(\pi)/(8)`, then solving for
`x` gives us:

`u+(\pi)/(8)=x` ( I just added
`(\pi)/(8)` on both sides.)

So recall
`0\le x<2\pi`.

Then
`0 \le u+(\pi)/(8)<2 \pi`.

Subtract
`(\pi)/(8)` on both sides:

`-(\pi)/(8)\le u <2 \pi-(\pi)/(8)`

Simplify:

`-(\pi)/(8)\le u <\pi (2-(1)/(8))`

`-(\pi)/(8)\le u<(15\pi)/(8)`

So we want to find solutions to:

`\tan(u)=(1)/(√(3))` with the condition:

`-(\pi)/(8)\le u<(15\pi)/(8)`

That's just at
`(\pi)/(6)` and
`(7\pi)/(6)`

So now adding
`(\pi)/(8)` to both gives us the solutions to:

`\tan(x-(\pi)/(8))=(1)/(√(3))` in the interval:

`0\le x<2\pi`.

The solutions we are looking for are:

`(\pi)/(6)+(\pi)/(8)` and
`(7\pi)/(6)+(\pi)/(8)`

Let's simplifying:

`((1)/(6)+(1)/(8))\pi` and
`((7)/(6)+(1)/(8))\pi`

`(7)/(24)\pi` and
`(31)/(24)\pi`

`(7\pi)/(24)` and
`(31\pi)/(24)`