By request:
General solutions to circular functions.
Many people just simply apply a formula (similar to the one provided in the Essentials 3/4 text), some knows where it comes from but most just mindlessly apply the formulas without knowing where it comes from and if the question gets tricky then they will get stuck. Hopefully this tutorial help you understand how we go about finding general solutions to circular functions in a more systematic fashion.
We will consider the general solutions to each of our 3 main circular functions,
,
and
. First we will look at the
function.
Example 1. Find the general solution to
 = \frac{1}{2})
First notice why does this question say find the GENERAL solution? This is because no domain is specified, if no domain is specified then there are infinitely many solutions to the above equation. [What this means is that if you sketch the graph of
)
and draw the horizontal line
, then there are infinitely many x values which gives a y value of
]
So here is how I would go about solving this question.
Let
 \cdots [1])
Now we have the equation
 = \frac{1}{2})
so let us find the 2 basic solutions to
then we will use
to use the 2 basic solutions to
to find the 2 basic solutions for
.
Solving
is quite trivial. This is an "exact value" question.
Now if you do not know how I solved the above equation, you need to review your circular function fundamentals ASAP.
Substituting the 2 basic
value into [1] yields:
, \frac{1}{3}\left(\frac{5\pi}{6}+\frac{3\pi}{4}\right))
Now the next step is finding the GENERAL solutions for x.
Look at the following graph of
The red line is the line
, as you can see, since a domain is not specified, it crosses the sin graph infinitely many times.
Now why did we find TWO basic solutions and not just one? As you can see the purple lines represent the solutions obtained from
and to get the other purple line solutions we simply have to add and subtract periods away from our basic solution of
.
But as you can see from the graph, no matter how many periods we add or subtract we will never end up on the green lines and this is what the other basic is for!
If we add and subtract periods away from
then we reach all the other green solutions.
So what is the period of the graph? Well it's
, again go back and review your fundamentals if you don't know how to calculate periods.
So our general solution is:
where
Now notice some of you might go, "wait what? Didn't you say we must SUBTRACT periods as well as adding them?"
This is another common mistake students often make, look at my definition of
in my answer. I said
is an INTEGER which means
ITSELF can take on negative values, eg, n = ...-3,-2,-1,0,1,2,3...
So for example say n = 1
Then we have
So here we are adding periods.
But if n = -1 then we have:
, \frac{19\pi}{36} + \frac{2\pi}{3} \times (-1))
Which is equivalent to:
So here we are subtracting periods.
So that is why we don't write our solution as:
where
Because the subtracting periods is already taken into account due to the restriction on
However some of you like to have the
in the middle and another way of writing the answer is this: (Note the difference!)
where
Why do we need
in the middle here? This is because n is now an element of NATURAL numbers or 0, which means n = 0, 1, 2, 3 ...
So the 'subtracting' periods is NOT taken into account from our restriction on n, that is why we need to put
in the middle since we need to 'manually' take into consideration ADDING and SUBTRACTING periods.
Both way of presenting the answer is fine, pick one and stick to it
Example 2.Find the general solution to
 = \frac{1}{2})
NO DIFFERENCE, APPROACH THIS QUESTION THE EXACT SAME WAY AS EXAMPLE 1, TRY IT YOURSELF!
Example 3.Find the general solution to
 =\sqrt{3})
Now the tan function is a tiny bit different in that we only need to find ONE basic solution and not TWO. The rest of the principles of adding and subtracting periods is all the same.
Here is how I would solve this question:
Let us first sketch the graph of
)
below:
The red line is the line
and the purple lines are the solutions to the equation.
As you can see from the graph, by finding any value of the x value that corresponds to the purple line and then adding and subtracting periods from that x value we get, we will be able to find all the solutions! So we don't need to solve for TWO basic solutions, ONE will be enough! (You might ask why do we need to solve for just one basic solution, graphically I have explained it, but algebraically this is because
)
is a one to one function while sin and cos are not. You don't really need to know this though)
So let us solve it!
 = \sqrt{3})
Thus the general solution is:
where
(Note the period of
)
is
and not
)
OR another way of writing it is:
where
Well that's it folks, enjoy and post any questions if you don't understand!
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