Lesson Explainer: Solving a Trigonometric Equation | Nagwa Lesson Explainer: Solving a Trigonometric Equation | Nagwa

Lesson Explainer: Solving a Trigonometric Equation Mathematics

In this explainer, we will learn how to solve a trigonometric equation using factoring or squaring.

Definition:

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 sin𝜃=𝑘, cos2𝜃=𝑘, and tan(𝜃30)=𝑘 using either a graph or a CAST diagram. We should also recall some key properties relating to the sine, cosine, and tangent functions:

  • sinsin𝜃=(180𝜃)
  • coscos𝜃=(360𝜃)
  • tantan𝜃=(180+𝜃)

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 sin𝑥=0.5 in the interval 0𝑥360, we draw a horizontal line at 𝑦=0.5 across the graph of 𝑦=𝑥sin and determine that this line intersects the graph twice in the given interval. We can therefore infer that the equation sin𝑥=0.5 has two solutions in the interval 0𝑥360.

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 𝜃, tansincos𝜃𝜃𝜃.

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

Fill in the blank: If 0𝑥360, then the number of solutions of the equation 4𝑥=𝑥sintan is .

Answer

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 tansincos𝑥𝑥𝑥. Making this substitution on the right-hand side of the equation, we have 4𝑥=𝑥𝑥4𝑥𝑥𝑥=0.sinsincossinsincos

The temptation at this stage may be to divide the equation through by the shared factor of sin𝑥. 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 sin𝑥 gives sincos𝑥41𝑥=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 sin𝑥=0. Recalling the graph of the sine function, we see that sin𝑥=0 three times in the interval 0𝑥360.

Incidentally, these values of 𝑥 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 sin𝑥; 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 41𝑥=0,cos which requires some further manipulation. Multiplying by cos𝑥 gives 4𝑥1=04𝑥=1𝑥=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 𝑥 in the interval 0𝑥360 for which cos𝑥=14:

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

There are 5 solutions to the equation 4𝑥=𝑥sintan in the interval 0𝑥360.

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 tantan𝜃+𝜃=0, where 0𝜃<180.

Answer

Upon inspection, we may identify immediately that this is a quadratic equation in tan𝜃. Alternatively, we may find it helpful to consider the substitution 𝑥=𝜃tan. We will consider the working out both with and without this substitution in parallel. LettanThenConsidertheequationtantanFactoringbygivesFactoringbytangivestantanSettingeachfactorequaltowehaveSettingeachfactorequaltowehaveortanortanReversingthesubstitution,weobtaintanortan𝑥=𝜃.𝑥+𝑥=0.𝜃+𝜃=0.𝑥𝜃𝑥(𝑥+1)=0.𝜃(𝜃+1)=0.0,0,𝑥=0𝑥+1=0.𝜃=0𝜃+1=0.𝜃=0𝜃+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 tan𝜃=0. Recalling the graph of the tangent function, we see that the only solution to this equation in the interval 0𝜃<180 is 𝜃=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 𝜃 that are strictly less than 180.

Our second equation can be rearranged to tan𝜃=1. Applying the inverse tangent function, we obtain 𝜃=(1)=45.tan

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

This value is in the required interval and so is a valid solution for 𝜃. 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 tantan𝜃+𝜃=0, where 0𝜃<180, is {0,135}.

Note that in the preceding example it was important that we factored by tan𝜃, rather than divided by it. Had we divided by tan𝜃, we would have lost one of the solutions to the original equation. This is because it is possible for tan𝜃 to equal 0 in the interval in which we were searching for solutions, and division by 0 is undefined. Indeed, tan𝜃=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 0𝜃360, 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 22𝜃+2𝜃=0coscos, given that 0𝜃360.

Answer

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 2𝜃cos: 2𝜃2𝜃+1=0.coscos

Hence, either 2𝜃=0cos or 2𝜃+1=0cos.

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

Solving the second equation is a little more complex. Rearranging gives 2𝜃=1𝜃=12.coscos

By multiplying both the numerator and denominator of the quotient on the right-hand side by 2, this can be rewritten as cos𝜃=22.

We should recall that cos45=22. However, we are looking for the values of 𝜃 such that cos𝜃=22. Using a CAST diagram, we see that cos𝜃 is negative in the second and third quadrants, and using symmetry we find that cos𝜃=22 when 𝜃=18045 and 𝜃=180+45.

This gives two further solutions: 135 and 225.

The set of values for which 22𝜃+2𝜃=0coscos in the interval 0𝜃360 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 3𝜃2𝜃𝜃=0sinsincos, where 0𝜃360. Give the answer to the nearest minute.

Answer

We begin by factoring by the shared factor of sin𝜃: sinsincos𝜃(3𝜃2𝜃)=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 sin𝜃=0.

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

Setting the parentheses equal to 0 yields the equation 3𝜃2𝜃=0sincos, which can be rearranged to sincos𝜃=23𝜃.

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

Recalling the identity tansincos𝜃𝜃𝜃, we can now express the equation in terms of a single trigonometric function: tan𝜃=23.

Solving for the principal value of 𝜃, correct to the nearest minute, gives 𝜃=23=33.6903341.tan

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

Values of tan𝜃 are also positive in the third quadrant, so we find the second value by adding 180 to our principal value: 𝜃=180+3341=21341.

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 {0,3341,180,21341}.

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 sincos𝜃+𝜃, which can be simplified by recalling the Pythagorean identity.

Definition: The Pythagorean Identity

For all values of 𝜃, sincos𝜃+𝜃1.

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

As squaring and finding the square root are not one-to-one operations, we have “created” an extra solution to this equation, 𝑥=3. This is known as an extraneous solution and is not correct given that our starting point was the single value 𝑥=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 4𝜃4𝜃=3sincos, where 0<𝜃360. Be careful to remove any extraneous solutions. Give your answer to two decimal places.

Answer

The question advises that we approach this problem by first squaring both sides of the equation. Doing so, and then simplifying, we obtain (4𝜃4𝜃)=316𝜃32𝜃𝜃+16𝜃=3.sincossinsincoscos

We then recall the Pythagorean identity sincos𝜃+𝜃1, which allows us to simplify further: 16𝜃+𝜃32𝜃𝜃=31632𝜃𝜃=332𝜃𝜃=13𝜃𝜃=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 sin𝜃 and cos𝜃, 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 sin𝜃 in terms of cos𝜃: 4𝜃4𝜃=34𝜃=3+4𝜃𝜃=3+4𝜃4.sincossincossincos

We can now substitute this expression for sin𝜃 into our second equation: 3+4𝜃4𝜃=1332.coscos

Distributing the parentheses, multiplying through by 32, and collecting all the terms on one side of the equation, we have 34𝜃+𝜃=133283𝜃+32𝜃=1332𝜃+83𝜃13=0.coscoscoscoscoscos

We now have a quadratic equation in cos𝜃, which we can solve by applying the quadratic formula. The coefficient of cos𝜃 is 32, the coefficient of cos𝜃 is 83, and the constant term is 13. Substituting these values into the quadratic formula gives cos𝜃=83±83(4×32×13)2×32.

Simplifying, we arrive at cos𝜃=83±185664, which we can simplify further by simplifying the radical 1856 and then canceling a factor of 8: cos𝜃=83±82964=3±298.

We now have two equations to solve. Taking the positive root first gives coscos𝜃=3+298𝜃=3+298=62.82962.83.

Using the symmetry of the cosine function, we have a second possible solution: 𝜃=36062.829=297.170297.17.

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 coscos𝜃=3298𝜃=3298=152.829152.83.

Again, using the symmetry of the cosine function, our second possible solution is 𝜃=360152.829=207.170207.17.

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 “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 3(1.732).

For our first value of 62.83: 462.83462.83=1.732,sincos and so 62.83 is a valid solution.

For the second value of 152.83: 4152.834152.83=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 4𝜃4𝜃=3sincos in the given interval, which to two decimal places are 𝜃=62.83 and 𝜃=207.17.

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 𝜃, 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 tansincos𝜃𝜃𝜃 and sincos𝜃+𝜃1 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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