In this explainer, we will learn how to solve a trigonometric equation using factoring or squaring.
A trigonometric equation is an equation which involves any of the three trigonometric functions sine, cosine, or tangent.
Before we embark on these new methods, we should already be familiar with solving simple trigonometric equations such as , , and using either a graph or a CAST diagram. We should also recall some key properties relating to the sine, cosine, and tangent functions:
Before solving a trigonometric equation, it is often helpful to consider how many solutions we expect the equation to have, as this can provide a means of sense-checking our answer. We can identify the number of times a trigonometric function is equal to a particular value in a given interval by drawing a horizontal line across its graph at this value and then counting the number of times this line intersects the graph. For example, if we want to determine the number of solutions to the equation in the interval , we draw a horizontal line at across the graph of and determine that this line intersects the graph twice in the given interval. We can therefore infer that the equation has two solutions in the interval .
Let us now look at a more complex example of this type of problem. We will also require one of the key trigonometric identities that defines the relationship between the three trigonometric functions. We recall this identity below.
Definition: The Tangent Identity
For all values of ,
Example 1: Identifying the Number of Solutions to a Trigonometric Equation
If , then the number of solutions of the equation is .
This equation involves two trigonometric ratios: sine and tangent. However, we can express the tangent function in terms of the sine and cosine functions by recalling the identity . Making this substitution on the right-hand side of the equation, we have
The temptation at this stage may be to divide the equation through by the shared factor of . However, doing so could potentially result in the loss of some solutions if the factor we divide by is equal to 0. We know that division by 0 is undefined, and so if there is any risk that the term we divide by is equal to 0, we must factor the equation by this term instead. Factoring by gives
We now have a product equal to 0. The only way a product can equal 0 is if at least one of the factors is itself equal to 0. To find all possible solutions, we set each factor equal to 0 and solve the resulting equations. In this question, we are not actually looking to solve each equation fully, but rather to determine the number of solutions in the specified interval.
Setting the first factor equal to 0, we have the equation . Recalling the graph of the sine function, we see that three times in the interval .
Incidentally, these values of are , , and , but the values are not required for this problem. This is why it is key that we factor rather than divide through by the shared factor of ; if we had divided, then we would have lost these three valid solutions to the equation.
Setting the second factor equal to 0 yields the equation which requires some further manipulation. Multiplying by gives
Recalling the graph of the cosine function and drawing a horizontal line across the graph at , we find that there are two values of in the interval for which :
While we cannot read the exact values of from our graph, they are clearly not the same as the values for which , and therefore all our solutions are unique.
There are 5 solutions to the equation in the interval .
Let us now consider a detailed example of how we can find all the solutions of a more complex trigonometric equation by factoring.
Example 2: Finding the Solutions to a Trigonometric Equation over a Specified Range Using Factoring
Find the set of values satisfying , where .
Upon inspection, we may identify immediately that this is a quadratic equation in . Alternatively, we may find it helpful to consider the substitution . We will consider the working out both with and without this substitution in parallel.
We see then that we arrive at the same pair of equations. The choice of whether to make a formal substitution is personal, and this step can be omitted as familiarity with these types of problems increases. We now have two equations to solve.
Our first equation is . Recalling the graph of the tangent function, we see that the only solution to this equation in the interval is . We should also recognize that without sketching the graph, as this is one of the key angles for which we should commit the values of the three trigonometric ratios to memory. Note that whilst , is not a solution as the interval contains only values of that are strictly less than .
Our second equation can be rearranged to . Applying the inverse tangent function, we obtain
This value, or principal argument, is outside the required interval for . Recalling the periodocity of the tangent function, we find a second solution to the equation by adding :
This value is in the required interval and so is a valid solution for . Adding any further multiples of would take us beyond the upper end of the interval, so we have found all solutions.
The set of values satisfying , where , is .
Note that in the preceding example it was important that we factored by , rather than divided by it. Had we divided by , we would have lost one of the solutions to the original equation. This is because it is possible for to equal 0 in the interval in which we were searching for solutions, and division by 0 is undefined. Indeed, was one of the equations we were subsequently required to solve.
We should always take careful note of the interval in which we are looking for solutions. It is common for the interval to be , but, as we saw in the previous example, this is not always the case. Additional values may well be valid solutions to the equation, but if they are out of the specified interval, then they are not correct in the context of the problem.
Let us now consider a slightly more complex problem involving a different trigonometric function.
Example 3: Finding the Solutions to a Trigonometric Equation over a Specified Range Using Factoring
Find the set of values satisfying , given that .
This equation involves a single trigonometric function, the cosine function, but it involves a term of order 2 and is therefore a quadratic equation. As it has no constant term (or the constant term is equal to 0), it can be solved by factoring by :
Hence, either or .
For the first equation, we divide through by 2 to give . Considering the graph of the cosine function, we see that there are two solutions to this equation in the required interval: and .
Solving the second equation is a little more complex. Rearranging gives
By multiplying both the numerator and denominator of the quotient on the right-hand side by , this can be rewritten as
We should recall that . However, we are looking for the values of such that . Using a CAST diagram, we see that is negative in the second and third quadrants, and using symmetry we find that when and .
This gives two further solutions: and .
The set of values for which in the interval is .
The previous two examples have consisted of solving quadratic equations in one trigonometric function only. Let us now consider an example in which we encounter an equation involving two trigonometric functions.
Example 4: Solving a Trigonometric Equation Involving Sine and Cosine by Factoring
Find the set of values satisfying , where . Give the answer to the nearest minute.
We begin by factoring by the shared factor of :
As we now have a product equal to 0, we can solve by setting each factor equal to 0 and then solving the resulting equations. Setting the first factor equal to 0 gives the equation .
Recalling the graph of the sine function, we find that there are three values of in the required interval: , , and .
Setting the parentheses equal to 0 yields the equation , which can be rearranged to
To express this equation in terms of one trigonometric function only, we can divide through by . Dividing through by a trigonometric term can be risky as we must ensure we do not lose any solutions, which can occur if the term we divide by is equal to 0. However, if were equal to 0, this would lead to , and as there are no values of for which sine and cosine are both zero, we know that is not equal to 0 for any of our solutions. Hence, we can be confident that dividing through by will not result in the loss of any solutions, and we obtain
Recalling the identity , we can now express the equation in terms of a single trigonometric function:
Solving for the principal value of , correct to the nearest minute, gives
To find any other values in the required interval, we consider the CAST diagram:
Values of are also positive in the third quadrant, so we find the second value by adding to our principal value:
We could also find this second value by recalling the periodicity of the tangent function, which would again require us to add . There are no further values in the required interval, so we have found all possible solutions.
The set of values satisfying the given equation, each to the nearest minute, is .
Another approach we can take to solving some specific types of trigonometric equations is to square both sides of the equation. In some cases, this can create expressions of the form , which can be simplified by recalling the Pythagorean identity.
Definition: The Pythagorean Identity
For all values of ,
Squaring both sides of an equation can be risky if we do not take great care. For example, consider the simple equation . Squaring both sides gives , and then solving this equation by finding the square root gives
As squaring and finding the square root are not one-to-one operations, we have “created” an extra solution to this equation, . This is known as an extraneous solution and is not correct given that our starting point was the single value . If we need to solve a trigonometric equation by squaring, we must subsequently check all our solutions in the original equation to ensure we have not obtained any extraneous values.
Example 5: Finding the Solutions to a Trigonometric Equation over a Specified Range by Squaring and Recognizing Extraneous Solutions
By first squaring both sides, or otherwise, solve the equation , where . Be careful to remove any extraneous solutions. Give your answer to two decimal places.
The question advises that we approach this problem by first squaring both sides of the equation. Doing so, and then simplifying, we obtain
We then recall the Pythagorean identity , which allows us to simplify further:
This equation is in terms of both the sine and cosine functions. Combining this with our original equation, we now have two equations in the two variables and , and so this system of equations can be solved simultaneously. In order to obtain an equation in one function only, we first rearrange the original equation to express in terms of :
We can now substitute this expression for into our second equation:
Distributing the parentheses, multiplying through by 32, and collecting all the terms on one side of the equation, we have
We now have a quadratic equation in , which we can solve by applying the quadratic formula. The coefficient of is 32, the coefficient of is , and the constant term is . Substituting these values into the quadratic formula gives
Simplifying, we arrive at which we can simplify further by simplifying the radical and then canceling a factor of 8:
We now have two equations to solve. Taking the positive root first gives
Using the symmetry of the cosine function, we have a second possible solution:
We refer to this as a “possible” solution as we must later check all our values in the original equation to determine whether any are extraneous.
Taking the negative root, we have
Again, using the symmetry of the cosine function, our second possible solution is
We have found four possible solutions to the given equation, the set of values .
Finally, we must check the validity of each of these “solutions” in the original equation. We substitute each value in turn into the left-hand side of the equation and determine whether the value we obtain is equal to .
For our first value of : and so is a valid solution.
For the second value of : and so is in fact an extraneous solution. In the same way, we find that our third value of is a valid solution, but the fourth value of is not.
Hence there are two solutions to the equation in the given interval, which to two decimal places are and .
The previous example highlights the importance of checking all our “solutions” in the original equation, if the method we used to solve it involved squaring. Had we omitted this step, we would have offered four, rather than two, values for , two of which would have been invalid.
We have seen various methods for solving more complex trigonometric equations including factoring, squaring, and using the trigonometric identities. In some instances, it will be possible to apply more than one method, and so the technique we choose will depend on the type and complexity of the equation given.
Let us finish by recapping some key points.
- Some trigonometric equations can be solved by factoring. It is extremely important that any common factors are factored by, not divided by, to avoid the potential loss of solutions if these factors are equal to 0.
- Some trigonometric equations can be solved by squaring both sides. Whenever this approach is used, care must be taken to avoid creating any extraneous solutions.
- The two key trigonometric identities and can be useful in simplifying trigonometric equations.
- The graphs of trigonometric functions, their properties, and the CAST diagram can be used to determine additional solutions following determination of a principal value.