Lesson Explainer: Solving a Trigonometric Equation Mathematics

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 sinπœƒ=π‘˜, cos2πœƒ=π‘˜, and tan(πœƒβˆ’30)=π‘˜ using either a graph or a CAST diagram. We should also recall some key properties relating to the sine, cosine, and tangent functions:

  • sinsinπœƒ=(180βˆ’πœƒ)∘∘∘
  • coscosπœƒ=(360βˆ’πœƒ)∘∘∘
  • tantanπœƒ=(180+πœƒ)∘∘∘

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 sinπ‘₯=0.5 in the interval 0≀π‘₯≀360∘∘, we draw a horizontal line at 𝑦=0.5 across the graph of 𝑦=π‘₯sin and determine that this line intersects the graph twice in the given interval. We can therefore infer that the equation sinπ‘₯=0.5 has two solutions in the interval 0≀π‘₯≀360∘∘.

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 πœƒ, tansincosπœƒβ‰‘πœƒπœƒ.

Example 1: Identifying the Number of Solutions to a Trigonometric Equation

If 0≀π‘₯≀360∘∘, then the number of solutions of the equation 4π‘₯=π‘₯sintan 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 tansincosπ‘₯≑π‘₯π‘₯. Making this substitution on the right-hand side of the equation, we have 4π‘₯=π‘₯π‘₯4π‘₯βˆ’π‘₯π‘₯=0.sinsincossinsincos

The temptation at this stage may be to divide the equation through by the shared factor of sinπ‘₯. 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 sinπ‘₯ gives sincosπ‘₯ο€Ό4βˆ’1π‘₯=0.

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 sinπ‘₯=0. Recalling the graph of the sine function, we see that sinπ‘₯=0 three times in the interval 0≀π‘₯≀360∘∘.

Incidentally, these values of π‘₯ are 0∘, 180∘, and 360∘, 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 sinπ‘₯; 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 4βˆ’1π‘₯=0,cos which requires some further manipulation. Multiplying by cosπ‘₯ gives 4π‘₯βˆ’1=04π‘₯=1π‘₯=14.coscoscos

Recalling the graph of the cosine function and drawing a horizontal line across the graph at 𝑦=14, we find that there are two values of π‘₯ in the interval 0≀π‘₯≀360∘∘ for which cosπ‘₯=14:

While we cannot read the exact values of π‘₯ from our graph, they are clearly not the same as the values for which sinπ‘₯=0, and therefore all our solutions are unique.

There are 5 solutions to the equation 4π‘₯=π‘₯sintan in the interval 0≀π‘₯≀360∘∘.

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 tantanοŠ¨πœƒ+πœƒ=0, where 0β‰€πœƒ<180∘∘.


Upon inspection, we may identify immediately that this is a quadratic equation in tanπœƒ. Alternatively, we may find it helpful to consider the substitution π‘₯=πœƒtan. We will consider the working out both with and without this substitution in parallel.LettanThenConsidertheequationtantanFactoringbygivesFactoringbytangivestantanSettingeachfactorequaltowehaveSettingeachfactorequaltowehaveortanortanReversingthesubstitution,weobtaintanortanπ‘₯=πœƒ.π‘₯+π‘₯=0.πœƒ+πœƒ=0.π‘₯πœƒπ‘₯(π‘₯+1)=0.πœƒ(πœƒ+1)=0.0,0,π‘₯=0π‘₯+1=0.πœƒ=0πœƒ+1=0.πœƒ=0πœƒ+1=0.

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 tanπœƒ=0. Recalling the graph of the tangent function, we see that the only solution to this equation in the interval 0β‰€πœƒ<180∘∘ is πœƒ=0∘. We should also recognize that tan0=0∘ 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 tan180=0∘, 180∘ is not a solution as the interval contains only values of πœƒ that are strictly less than 180∘.

Our second equation can be rearranged to tanπœƒ=βˆ’1. Applying the inverse tangent function, we obtain πœƒ=(βˆ’1)=βˆ’45.tan∘

This value, or principal argument, is outside the required interval for πœƒ. Recalling the periodocity of the tangent function, we find a second solution to the equation by adding 180∘: πœƒ=βˆ’45+180=135.∘∘∘

This value is in the required interval and so is a valid solution for πœƒ. Adding any further multiples of 180∘ would take us beyond the upper end of the interval, so we have found all solutions.

The set of values satisfying tantanοŠ¨πœƒ+πœƒ=0, where 0β‰€πœƒ<180∘∘, is {0,135}∘.

Note that in the preceding example it was important that we factored by tanπœƒ, rather than divided by it. Had we divided by tanπœƒ, we would have lost one of the solutions to the original equation. This is because it is possible for tanπœƒ to equal 0 in the interval in which we were searching for solutions, and division by 0 is undefined. Indeed, tanπœƒ=0 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 0β‰€πœƒβ‰€360∘∘, 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 2√2πœƒ+2πœƒ=0coscos, given that 0β‰€πœƒβ‰€360∘∘.


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 2πœƒcos: 2πœƒο€»βˆš2πœƒ+1=0.coscos

Hence, either 2πœƒ=0cos or √2πœƒ+1=0cos.

For the first equation, we divide through by 2 to give cosπœƒ=0. Considering the graph of the cosine function, we see that there are two solutions to this equation in the required interval: πœƒ=90∘ and πœƒ=270∘.

Solving the second equation is a little more complex. Rearranging gives √2πœƒ=βˆ’1πœƒ=βˆ’1√2.coscos

By multiplying both the numerator and denominator of the quotient on the right-hand side by √2, this can be rewritten as cosπœƒ=βˆ’βˆš22.

We should recall that cos45=√22∘. However, we are looking for the values of πœƒ such that cosπœƒ=βˆ’βˆš22. Using a CAST diagram, we see that cosπœƒ is negative in the second and third quadrants, and using symmetry we find that cosπœƒ=βˆ’βˆš22 when πœƒ=180βˆ’45∘∘ and πœƒ=180+45∘∘.

This gives two further solutions: 135∘ and 225∘.

The set of values for which 2√2πœƒ+2πœƒ=0coscos in the interval 0β‰€πœƒβ‰€360∘∘ is {90,135,225,270}∘∘∘∘.

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 3πœƒβˆ’2πœƒπœƒ=0sinsincos, where 0β‰€πœƒβ‰€360∘∘. Give the answer to the nearest minute.


We begin by factoring by the shared factor of sinπœƒ: sinsincosπœƒ(3πœƒβˆ’2πœƒ)=0.

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 sinπœƒ=0.

Recalling the graph of the sine function, we find that there are three values of πœƒ in the required interval: 0∘, 180∘, and 360∘.

Setting the parentheses equal to 0 yields the equation 3πœƒβˆ’2πœƒ=0sincos, which can be rearranged to sincosπœƒ=23πœƒ.

To express this equation in terms of one trigonometric function only, we can divide through by cosπœƒ. 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 cosπœƒ were equal to 0, this would lead to sinπœƒ=0, and as there are no values of πœƒ for which sine and cosine are both zero, we know that cosπœƒ is not equal to 0 for any of our solutions. Hence, we can be confident that dividing through by cosπœƒ will not result in the loss of any solutions, and we obtain sincosπœƒπœƒ=23.

Recalling the identity tansincosπœƒβ‰‘πœƒπœƒ, we can now express the equation in terms of a single trigonometric function: tanπœƒ=23.

Solving for the principal value of πœƒ, correct to the nearest minute, gives πœƒ=ο€Ό23=33.690β€¦β‰ˆ3341β€².tan∘∘

To find any other values in the required interval, we consider the CAST diagram:

Values of tanπœƒ are also positive in the third quadrant, so we find the second value by adding 180∘ to our principal value: πœƒ=180+3341β€²=21341β€².∘∘∘

We could also find this second value by recalling the periodicity of the tangent function, which would again require us to add 180∘. 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 {0,3341β€²,180,21341β€²}∘∘∘∘.

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 sincosοŠ¨οŠ¨πœƒ+πœƒ, which can be simplified by recalling the Pythagorean identity.

Definition: The Pythagorean Identity

For all values of πœƒ, sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1.

Squaring both sides of an equation can be risky if we do not take great care. For example, consider the simple equation π‘₯=3. Squaring both sides gives π‘₯=9, and then solving this equation by finding the square root gives π‘₯=±√9=Β±3.

As squaring and finding the square root are not one-to-one operations, we have β€œcreated” an extra solution to this equation, π‘₯=βˆ’3. This is known as an extraneous solution and is not correct given that our starting point was the single value π‘₯=3. 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 4πœƒβˆ’4πœƒ=√3sincos, where 0<πœƒβ‰€360∘∘. 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 (4πœƒβˆ’4πœƒ)=ο€»βˆš316πœƒβˆ’32πœƒπœƒ+16πœƒ=3.sincossinsincoscos

We then recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1, which allows us to simplify further: 16ο€Ίπœƒ+πœƒο†βˆ’32πœƒπœƒ=316βˆ’32πœƒπœƒ=3βˆ’32πœƒπœƒ=βˆ’13πœƒπœƒ=1332.sincossincossincossincossincos

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 sinπœƒ and cosπœƒ, 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 sinπœƒ in terms of cosπœƒ: 4πœƒβˆ’4πœƒ=√34πœƒ=√3+4πœƒπœƒ=√3+4πœƒ4.sincossincossincos

We can now substitute this expression for sinπœƒ into our second equation: ο€Ώβˆš3+4πœƒ4ο‹πœƒ=1332.coscos

Distributing the parentheses, multiplying through by 32, and collecting all the terms on one side of the equation, we have √34πœƒ+πœƒ=13328√3πœƒ+32πœƒ=1332πœƒ+8√3πœƒβˆ’13=0.coscoscoscoscoscos

We now have a quadratic equation in cosπœƒ, which we can solve by applying the quadratic formula. The coefficient of cosοŠ¨πœƒ is 32, the coefficient of cosπœƒ is 8√3, and the constant term is βˆ’13. Substituting these values into the quadratic formula gives cosπœƒ=βˆ’8√3Β±ο„žο€»8√3ο‡βˆ’(4Γ—32Γ—βˆ’13)2Γ—32.

Simplifying, we arrive at cosπœƒ=βˆ’8√3±√185664, which we can simplify further by simplifying the radical √1856 and then canceling a factor of 8: cosπœƒ=βˆ’8√3Β±8√2964=βˆ’βˆš3±√298.

We now have two equations to solve. Taking the positive root first gives coscosπœƒ=βˆ’βˆš3+√298πœƒ=ο€Ώβˆ’βˆš3+√298=62.829β€¦β‰ˆ62.83.

Using the symmetry of the cosine function, we have a second possible solution: πœƒ=360βˆ’62.829…=297.170β€¦β‰ˆ297.17.

We refer to this as a β€œpossible” 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 coscosπœƒ=βˆ’βˆš3βˆ’βˆš298πœƒ=ο€Ώβˆ’βˆš3βˆ’βˆš298=152.829β€¦β‰ˆ152.83.

Again, using the symmetry of the cosine function, our second possible solution is πœƒ=360βˆ’152.829…=207.170β€¦β‰ˆ207.17.

We have found four possible solutions to the given equation, the set of values {62.83,152.83,207.17,297.17}∘∘∘∘.

Finally, we must check the validity of each of these β€œsolutions” 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 √3(1.732…).

For our first value of 62.83∘: 462.83βˆ’462.83=1.732…,sincos∘∘ and so 62.83∘ is a valid solution.

For the second value of 152.83∘: 4152.83βˆ’4152.83=5.385…,sincos∘∘ and so 152.83∘ is in fact an extraneous solution. In the same way, we find that our third value of 207.17∘ is a valid solution, but the fourth value of 297.17∘ is not.

Hence there are two solutions to the equation 4πœƒβˆ’4πœƒ=√3sincos in the given interval, which to two decimal places are πœƒ=62.83∘ and πœƒ=207.17∘.

The previous example highlights the importance of checking all our β€œsolutions” 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 πœƒ, 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.

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 tansincosπœƒβ‰‘πœƒπœƒ and sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1 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.

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