Answer:
and ![\frac{31\pi}{24}](https://tex.z-dn.net/?f=%5Cfrac%7B31%5Cpi%7D%7B24%7D)
Step-by-step explanation:
![\sqrt{3} \tan(x-\frac{\pi}{8})-1=0](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5Ctan%28x-%5Cfrac%7B%5Cpi%7D%7B8%7D%29-1%3D0)
Let's first isolate the trig function.
Add 1 one on both sides:
![\sqrt{3} \tan(x-\frac{\pi}{8})=1](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5Ctan%28x-%5Cfrac%7B%5Cpi%7D%7B8%7D%29%3D1)
Divide both sides by
:
![\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}](https://tex.z-dn.net/?f=%5Ctan%28x-%5Cfrac%7B%5Cpi%7D%7B8%7D%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D)
Now recall
.
![\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%7D)
or
![\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%3D%5Cfrac%7B-%5Cfrac%7B1%7D%7B2%7D%7D%7B-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%7D)
The first ratio I have can be found using
in the first rotation of the unit circle.
The second ratio I have can be found using
you can see this is on the same line as the
so you could write
as
.
So this means the following:
![\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}](https://tex.z-dn.net/?f=%5Ctan%28x-%5Cfrac%7B%5Cpi%7D%7B8%7D%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D)
is true when ![x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi](https://tex.z-dn.net/?f=x-%5Cfrac%7B%5Cpi%7D%7B8%7D%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%2Bn%20%5Cpi)
where
is integer.
Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.
So now we have a linear equation to solve:
![x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi](https://tex.z-dn.net/?f=x-%5Cfrac%7B%5Cpi%7D%7B8%7D%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%2Bn%20%5Cpi)
Add
on both sides:
![x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%2B%5Cfrac%7B%5Cpi%7D%7B8%7D%2Bn%20%5Cpi)
Find common denominator between the first two terms on the right.
That is 24.
![x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B4%5Cpi%7D%7B24%7D%2B%5Cfrac%7B3%5Cpi%7D%7B24%7D%2Bn%20%5Cpi)
(So this is for all the solutions.)
Now I just notice that it said find all the solutions in the interval
.
So if
and we let
, then solving for
gives us:
( I just added
on both sides.)
So recall
.
Then
.
Subtract
on both sides:
![-\frac{\pi}{8}\le u](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cle%20u%20%3C2%20%5Cpi-%5Cfrac%7B%5Cpi%7D%7B8%7D)
Simplify:
![-\frac{\pi}{8}\le u](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cle%20u%20%3C%5Cpi%20%282-%5Cfrac%7B1%7D%7B8%7D%29)
![-\frac{\pi}{8}\le u](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cle%20u%3C%5Cfrac%7B15%5Cpi%7D%7B8%7D)
So we want to find solutions to:
with the condition:
![-\frac{\pi}{8}\le u](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cle%20u%3C%5Cfrac%7B15%5Cpi%7D%7B8%7D)
That's just at
and ![\frac{7\pi}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B7%5Cpi%7D%7B6%7D)
So now adding
to both gives us the solutions to:
in the interval:
.
The solutions we are looking for are:
and ![\frac{7\pi}{6}+\frac{\pi}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B7%5Cpi%7D%7B6%7D%2B%5Cfrac%7B%5Cpi%7D%7B8%7D)
Let's simplifying:
and ![(\frac{7}{6}+\frac{1}{8})\pi](https://tex.z-dn.net/?f=%28%5Cfrac%7B7%7D%7B6%7D%2B%5Cfrac%7B1%7D%7B8%7D%29%5Cpi)
and ![\frac{31}{24}\pi](https://tex.z-dn.net/?f=%5Cfrac%7B31%7D%7B24%7D%5Cpi)
and ![\frac{31\pi}{24}](https://tex.z-dn.net/?f=%5Cfrac%7B31%5Cpi%7D%7B24%7D)