This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic.
The next level of complexity comes when we need to solve a trinomial which contains trigonometric functions. It is much easier in this case to use
temporary variables .
Consider solving
Here we notice that
occurs twice in the equation, hence we let
and rewrite:
That should look rather more familiar. We can immediately write down the factorised form and the solutions:
Next we just substitute back for the temporary variable:
And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the
range of the trigonometric function. In that case you need to discard that solution. For example consider the same equation with cosines instead of tangents
Using the same method we find that
The second solution cannot be valid as cosine must lie between
and 1. We must, therefore, reject the second equation. Only solutions to the first equation will be valid.
More complex trigonometric equations
Here are two examples on the level of the hardest trigonometric equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything!
Solve
for
We note that
occurs twice in the equation. So, let
. Then we have
Note that with practice you may be able to leave out this step.
Factorising yields
We thus get
Both equations are valid (
i.e. lie in the range of cosine).
General solution:
Now we find the specific solutions in the interval
. Appropriate values of
yield
Solve for
in the interval
:
Factorising yields
which gives two equations
General solution:
Specific solution in the interval
:
Solving trigonometric equations
Find the general solution of each of the following equations. Give answers to one decimal place.
Find all solutions in the interval
Find the general solution of each of the following equations. Give answers to one decimal place.