Answer:
and 
Step-by-step explanation:

Let's first isolate the trig function.
Add 1 one on both sides:

Divide both sides by
:

Now recall
.

or

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:

is true when 
where
is integer.
Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.
So now we have a linear equation to solve:

Add
on both sides:

Find common denominator between the first two terms on the right.
That is 24.

(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:

Simplify:


So we want to find solutions to:
with the condition:

That's just at
and 
So now adding
to both gives us the solutions to:
in the interval:
.
The solutions we are looking for are:
and 
Let's simplifying:
and 
and 
and 