Answer:
[tex]\frac{7\pi}{24}[/tex] and [tex]\frac{31\pi}{24}[/tex]
Step-by-step explanation:
[tex]\sqrt{3} \tan(x-\frac{\pi}{8})-1=0[/tex]
Let's first isolate the trig function.
Add 1 one on both sides:
[tex]\sqrt{3} \tan(x-\frac{\pi}{8})=1[/tex]
Divide both sides by [tex]\sqrt{3}[/tex]:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex]
Now recall [tex]\tan(u)=\frac{\sin(u)}{\cos(u)}[/tex].
[tex]\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}[/tex]
or
[tex]\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}[/tex]
The first ratio I have can be found using [tex]\frac{\pi}{6}[/tex] in the first rotation of the unit circle.
The second ratio I have can be found using [tex]\frac{7\pi}{6}[/tex] you can see this is on the same line as the [tex]\frac{\pi}{6}[/tex] so you could write [tex]\frac{7\pi}{6}[/tex] as [tex]\frac{\pi}{6}+\pi[/tex].
So this means the following:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex]
is true when [tex]x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi[/tex]
where [tex]n[/tex] is integer.
Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.
So now we have a linear equation to solve:
[tex]x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi[/tex]
Add [tex]\frac{\pi}{8}[/tex] on both sides:
[tex]x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi[/tex]
Find common denominator between the first two terms on the right.
That is 24.
[tex]x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi[/tex]
[tex]x=\frac{7\pi}{24}+n \pi[/tex] (So this is for all the solutions.)
Now I just notice that it said find all the solutions in the interval [tex][0,2\pi)[/tex].
So if [tex]\sqrt{3} \tan(x-\frac{\pi}{8})-1=0[/tex] and we let [tex]u=x-\frac{\pi}{8}[/tex], then solving for [tex]x[/tex] gives us:
[tex]u+\frac{\pi}{8}=x[/tex] ( I just added [tex]\frac{\pi}{8}[/tex] on both sides.)
So recall [tex]0\le x<2\pi[/tex].
Then [tex]0 \le u+\frac{\pi}{8}<2 \pi[/tex].
Subtract [tex]\frac{\pi}{8}[/tex] on both sides:
[tex]-\frac{\pi}{8}\le u <2 \pi-\frac{\pi}{8}[/tex]
Simplify:
[tex]-\frac{\pi}{8}\le u <\pi (2-\frac{1}{8})[/tex]
[tex]-\frac{\pi}{8}\le u<\frac{15\pi}{8}[/tex]
So we want to find solutions to:
[tex]\tan(u)=\frac{1}{\sqrt{3}}[/tex] with the condition:
[tex]-\frac{\pi}{8}\le u<\frac{15\pi}{8}[/tex]
That's just at [tex]\frac{\pi}{6}[/tex] and [tex]\frac{7\pi}{6}[/tex]
So now adding [tex]\frac{\pi}{8}[/tex] to both gives us the solutions to:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex] in the interval:
[tex]0\le x<2\pi[/tex].
The solutions we are looking for are:
[tex]\frac{\pi}{6}+\frac{\pi}{8}[/tex] and [tex]\frac{7\pi}{6}+\frac{\pi}{8}[/tex]
Let's simplifying:
[tex](\frac{1}{6}+\frac{1}{8})\pi[/tex] and [tex](\frac{7}{6}+\frac{1}{8})\pi[/tex]
[tex]\frac{7}{24}\pi[/tex] and [tex]\frac{31}{24}\pi[/tex]
[tex]\frac{7\pi}{24}[/tex] and [tex]\frac{31\pi}{24}[/tex]
