Solving Trig. Equations

MathsDirect

We have seen that you can find trig ratios for any angle, by determining which acute angle it is equivalent to and then deciding whether the function should be positive or negative, depending on the quadrant. The results are summarized below.

Quadrant Acute Angle to use Sin Cos Tan

The implication of this is that there will be more than one solution to the equation

In degrees, we know that x could equal 30o, but there is also a positive value of sin in the second quadrant, where the rule is that you subtract the acute angle from 180. Therefore 150 is also a solution of our equation.

There are in fact an infinite number of solutions to the equation

Therefore, when ever you are asked to solve a trigonometric equation, you will need to find all the values in a given range.

For example, find all the solutions of the equation in the given range.

First find the first solution ( in the first quadrant).

This is a result that you should know (equivalent to 60o).

To find the second result, refer to the table above (which you should memorize). You can see that cos also has a positive value in the third quadrant, which you find by adding 180o

Therefore the second solution is

Can we stop here? Our range is to

so it is unlikely that we have found all of our solutions.

In general it is a good idea to continue finding solutions until you find a solution outside the range. This is easy to do, because the trig curves are periodic. This means that they keep repeating themselves exactly. If you look at the graph of cos x, you will see that it's period is

 

You can therefore find new solutions by adding (or subtracting) to your two existing solutions

This gives the following list of solutions

The last solution is outside the required range, so the answer to our question is

Exactly the same method can be used to find all the required solutions for an equation using sin and tan, with the exception that tan has a period of 180o.

On the next page there will be some more examples, covering the solutions of equations such as

 

Go to Next Page

Return to Trig Tutorial Contents

 

©2000 MathsDirect - All rights reserved   Terms&Conditions