Lesson Explainer: Simple Trigonometric Equations Mathematics

In this explainer, we will learn how to find angle measures given interval and function values.

A trigonometric equation is an equation that involves at least one of the following: a trigonometric function, such as sine, cosine, or tangent; a reciprocal trigonometric function, such as cosecant, secant, or cotangent; or an inverse of any of these. Some of the simpler examples of such equations can be solved without using a calculator; however, for the majority of these, it would be tedious to try and remember specific trigonometric function values. In these cases, we make use of the inverse of the functions alongside knowledge of the symmetry and periodicity of their graphs to calculate additional solutions.

Before we demonstrate how to use the symmetry of the graph of a trigonometric function to find all solutions in a given interval, we will first recall the exact values for the sine, cosine, and tangent of a number of special angles.

Consider a right isosceles triangle with two sides of length 1 cm, as shown in the figure.

Using the Pythagorean theorem, we can calculate the length of the hypotenuse to be 1+1=2cm. Then, by using trigonometric convention to label the sides with respect to the angle at the top of the diagram, we can calculate the exact value of sin(45).

sinopphypsin𝜃=45=12=22.

Using this isosceles triangle and the equilateral triangle below, you should be able to derive the following exact values for sine, cosine, and tangent (as shown in the table).

The exact values for the given values of 𝜃 are as follows.

𝜃030=𝜋645=𝜋460=𝜋390=𝜋2
sin𝜃01212=22321
cos𝜃13212=22120
tan𝜃013=3313Undefined

It is important that we are able to recall these exact values without feeling the need to reach for a calculator, and, as such, we should take time to familiarize ourselves with the tables given.

In the first example, we will demonstrate how to use the symmetry of the graph of the sine function, alongside the table of exact values, to find all solutions to a simple trigonometric equation.

Example 1: Identifying the General Solution to a Trigonometric Equation

What is the general solution of sin𝜃=22?

Answer

In order to find the general solution to a trigonometric equation, we begin by finding a particular solution. In this case, the table of exact trigonometric values can help.

For an angle 𝜃 given in radian measure, the exact values of the sine function are as follows.

𝜃0𝜋6𝜋4𝜋3𝜋2
sin𝜃01222321

We observe that sin𝜋4=22, so 𝜃=𝜋4 is a particular solution to the equation sin𝜃=22.

To find further solutions, we sketch the graph of 𝑦=𝜃sin as shown in the following figure.

The solutions to sin𝜃=22 are found by adding the line 𝑦=22 to the diagram. The values of 𝜃 at the points of intersection of this line with the sine curve are the solutions to the equation.

Since the sine curve has symmetry about 𝜋2 in the interval 0𝜃𝜋, the second solution is found by subtracting 𝜋4 from 𝜋: 𝜋𝜋4=3𝜋4.

Remember, the sine function is periodic with a period of 2𝜋 radians, so further solutions can be found by adding or subtracting 2𝜋 radians from the particular solutions.

In other words, the solution is 𝜋4+2𝑛𝜋, 3𝜋4+2𝑛𝜋, for integer values of 𝑛.

Alternatively, this can be represented using set notation as 𝜋4+2𝑛𝜋, 3𝜋4+2𝑛𝜋, where 𝑛.

In the previous example, we demonstrated how to interpret the symmetry of the graph of the sine function to find all solutions to an equation. Another powerful tool that can be used to extend the domain of the sine function is the unit circle. Remember, this is a circle centered at the origin with a radius of 1 unit. To use the unit circle to determine the sine of any angle 𝜃, we start at the point (1,0) and travel along the circumference of the circle in a counterclockwise direction until the angle that is formed between this point, the origin, and the positive 𝑥-axis is equal to 𝜃. Then, if this point has coordinates (𝑥,𝑦), sin𝜃 is equal to the value of 𝑦.

Notice that the value of the 𝑦-coordinate is positive in both the 1st and 2nd quadrants; hence, the value of sin𝜃 will also be positive in these quadrants. Also, since the unit circle has reflectional symmetry about the 𝑦-axis, we can see that sinsin𝜃=(180𝜃).

By continuing to move along the circumference of the unit circle, we can see that sinsin𝜃=(360+𝜃) for all values of 𝜃.

These results can be generalized as follows.

How To: Finding Solutions to Simple Equations Involving the Sine Function

If 𝜃=𝜃 is a solution to the equation sin𝜃=𝑐, for some constant 𝑐[1,1], a second solution is given by 𝜃=180𝜃. Then, the set of all solutions to sin𝜃=𝑐 is 𝜃=𝜃+2𝑛𝜋𝜃=𝜋𝜃+2𝑛𝜋,𝑛.andwhere

If 𝜃 is measured in degrees, the solutions are 𝜃=𝜃+360𝑛𝜃=180𝜃+360𝑛,𝑛.andwhere

While we might feel inclined to memorize these formulas, in practice, it can be much more effective to sketch the graph of the sine function or the unit circle to help us deduce the set of solutions to an equation involving the sine function. Let’s demonstrate this in our next example.

Example 2: Finding the Solutions to a Trigonometric Equation in a Specified Range

Find the set of values satisfying 4𝜃1=0sin, where 90𝜃360.

Answer

To solve this simple trigonometric equation, we begin by rearranging to make the sine function the subject: 4𝜃1=04𝜃=1𝜃=14.sinsinsin

Care must be taken with the next stage of calculations. We must take both the positive and negative square roots of 14. This will have the effect of creating a pair of equations in terms of sin𝜃: sinandsinsinandsin𝜃=14𝜃=14,𝜃=12𝜃=12.

Note that we might have also chosen to factor the expression 4𝜃1sin as (2𝜃1)(2𝜃+1)sinsin and then solve the equation (2𝜃1)(2𝜃+1)=0sinsin to achieve the same result.

Next, to solve the first equation, we recall the exact values for sin𝜃, where 𝜃 is measured in degrees, as follows.

𝜃030456090
sin𝜃01222321

Hence, a solution to the equation sin𝜃=12 is 𝜃=30. Note that this is outside of the specified interval 90𝜃360, so we will consider the symmetry of the unit circle to find further solutions.

We can see that the only solution in the given interval is 𝜃=18030=150.

Now, we will solve the second equation by, once again, using the symmetry of the unit circle. Since sin𝜃=𝑦 for all 𝜃 and the value of 𝑦 is negative in the 3rd and 4th quadrants, we add right triangles here as shown.

Hence, the next two solutions are 𝜃=180+30=210 and 𝜃=36030=330. At this stage, we might feel inclined to find further solutions by adding integer multiples of 360 to these, but doing so would result in values outside of the required interval.

The set of values satisfying 4𝜃1=0sin, where 90𝜃360, is {150,210,330}.

In the previous example, we demonstrated how to use the symmetry of the unit circle with information about exact values to solve trignometric equations where 𝜃 is not an acute angle. It is worth noting that in order to solve the second equation sin𝜃=12, we could have alternatively used the inverse sine function and a calculator to find 𝜃=12=30.sin

If we had chosen this method, we would have then needed to consider the regions in the unit circle where 𝜃<0 and applied the rules about symmetry and periodicity as normal.

In our next example, we will look at how to use the symmetry of the graph of the cosine function to solve a trignometric equation.

Example 3: Solving a Trigonometric Equation Involving a Shifted Angle in a Specified Range

Find the set of values satisfying cos(𝜃105)=12, where 0<𝜃<360.

Answer

In order to find the solutions to a trigonometric equation in a given interval, we begin by finding a particular solution. In this case, the table of exact trigonometric values can help.

We will first redefine the argument of the function by letting 𝛼=𝜃105 such that cosand𝛼=12𝜃=𝛼+105.

Amending the interval over which our solutions are valid by adding 105 gives 105<𝛼<465. Then, the exact values for cos𝛼, where 𝛼 is measured in degrees, are as follows.

𝛼030456090
cos𝛼13222120

We can see that cos𝛼=12; however, there are no values of 𝛼 given in the table such that cos𝛼=12. By sketching the graph of the cosine function and the lines 𝑦=12 and 𝑦=12, we can find the associated value of 𝛼.

The graph has rotational symmetry between 0𝛼180 about (90,0), so the first solution to cos𝛼=12 is 𝛼=18060=120.

Using the symmetry of the curve, the next solution is 𝛼=180+60=240.

There appears to be a further solution obtained by 𝛼=120+360=480; however, this is outside the interval 105<𝛼<465.

Hence, the solutions to cos𝛼=12 are 𝛼=120 and 𝛼=240.

Since we defined 𝜃=𝛼+105, the solutions to cos(𝜃105)=12 are 𝜃=120+105=225𝜃=240+105=345.and

Therefore, the solution set is {225,345}.

Note that an alternative technique to finding the particular solution to cos𝛼=12 is to use the inverse cosine function such that 𝛼=12=120.cos

At this stage, the remaining steps to find the other solutions are the same.

Remember, we can also use the symmetry of the unit circle to achieve the same result. Just as the value of the 𝑦-coordinate of the point of intersection of the terminal side of an angle with the unit circle tells us the value of sin𝜃, the value of the 𝑥-coordinate tells us the value of cos𝜃.

Since the value of the 𝑥-coordinate is positive in the 1st and 4th quadrants, by symmetry of the unit circle, coscos𝜃=(360𝜃).

Using the symmetry of the unit circle and periodicity of the cosine function, we can quote formulas for the general solution to equations involving this function.

How To: Solving Simple Equations Involving the Cosine Function

If 𝜃=𝜃 is a solution to the equation cos𝜃=𝑐, for some constant 𝑐[1,1], a second solution is given by 𝜃=360𝜃. Then, the set of all solutions to cos𝜃=𝑐 is 𝜃=𝜃+2𝑛𝜋𝜃=2𝜋𝜃+2𝑛𝜋,𝑛.andwhere

If 𝜃 is measured in degrees, the solutions are 𝜃=𝜃+360𝑛𝜃=360𝜃+360𝑛,𝑛.andwhere

In the previous example, we demonstrated how to solve a trigonometric equation where the argument of the trigonometric function has been transformed in some way. In such cases, we begin by redefining the argument and amending the interval over which we are solving, allowing us to use the symmetry of the standard trigonometric curves, before finally solving the resulting equations in 𝜃. This is, generally, a more sensible route to take than attempting to apply transformations to the curves of the trigonometric functions. This is summarized below.

How To: Solving a Simple Trigonometric Equation Involving a Transformed Argument

  1. Redefine the argument of the trigonometric function (e.g., let 𝑥=𝜃+30).
  2. Amend the given interval over which solutions are to be found using the same definition (e.g., 0𝜃360 means 30𝑥390).
  3. Find all solutions to the resulting equation over this new interval.
  4. Convert these solutions back into the original variable using the original definition (e.g., 𝜃=𝑥30).

We will now demonstrate how to use this technique to solve an equation involving a triple angle.

Example 4: Solving a Trigonometric Equation Involving a Triple Angle in a Specified Range

Find the set of values satisfying sin3𝑥=1, where 0𝑥2𝜋.

Answer

To solve sin3𝑥=1, we will begin by redefining the argument. Letting 𝜃=3𝑥 so that 𝑥=𝜃3, the equation becomes sin𝜃=1, where 0𝜃32𝜋. To simplify this interval, we multiply by 3 to get 0𝜃6𝜋.

Now, we can solve the equation sin𝜃=1 over this new interval. For an angle 𝜃 given in radian measure, the exact values of the sine function are as follows.

𝜃0𝜋6𝜋4𝜋3𝜋2
sin𝜃01222321

Hence, a solution to the equation sin𝜃=1 is 𝜃=𝜋2. In order to find further solutions, we will sketch the graph of 𝑦=𝜃sin over the interval 0𝜃6𝜋, remembering that the sine function is periodic with a period of 2𝜋 radians.

We can see that the solutions occur every 2𝜋 radians, so our additional solutions are 𝜃=5𝜋2 and 𝜃=9𝜋2.

Hence, the solution set of sin𝜃=1 over the required interval is 𝜋2,5𝜋2,9𝜋2.

Since 𝑥=𝜃3, we can find the solution set of sin3𝑥=1 by dividing each of these values by 3. The set of solutions is then 𝜋6,5𝜋6,3𝜋2.

We will now demonstrate how to apply this process to solve equations involving the tangent function.

Example 5: Solving a Trigonometric Equation Involving a Shifted Double Angle in a Specified Range

Find the set of values satisfying tan2𝑥+𝜋5=1, where 0𝑥2𝜋.

Answer

To solve this equation, we will begin by redefining the argument to allow us to use the symmetry of the tangent function. Let 𝜃=2𝑥+𝜋5, so tan𝜃=1 over 𝜋5𝜃21𝜋5.

Then, we can use the table of exact values and knowledge about the periodicity of the tangent function to solve this new equation.

Remember, for 𝜃 measured in radians, the exact values of tan𝜃 are as follows.

𝜃0𝜋6𝜋4𝜋3𝜋2
tan𝜃013=3313Undefined

We can see that tan𝜋4=1, so let’s use this to find the value of 𝜃, where tan𝜃=1. The graph of 𝑦=𝜃tan is sketched over the interval 0𝜃21𝜋5 below.

We can see that the graph of the tangent function has rotational symmetry about (𝑛𝜋,0), where 𝑛. Hence, the first solution to tan𝜃=1 is 𝜃=𝜋𝜋4=3𝜋4. Similarly, since the function is periodic with a period of 𝜋 radians, the remaining solutions are found by adding multiples of 𝜋 to this value: 𝜃=3𝜋4+𝜋=7𝜋4,𝜃=3𝜋4+2𝜋=11𝜋4,𝜃=3𝜋4+3𝜋=15𝜋4.

We now have four solutions to tan𝜃=1 over the required interval. Since we defined 𝜃=2𝑥+𝜋5, we can find the value of 𝑥 by setting 𝜃=2𝑥+𝜋5 in each case and solving for 𝑥. In the case of the first value of 𝜃, this gives 2𝑥+𝜋5=3𝜋42𝑥=11𝜋20𝑥=11𝜋40.

In a similar way, the remaining values of 𝑥 are 31𝜋40,51𝜋40, and 71𝜋40. Hence, the set of values satisfying tan2𝑥+𝜋5=1, where 0𝑥2𝜋, is 11𝜋40,31𝜋40,51𝜋40,71𝜋40.

In the previous example, we considered the periodicity of the tangent function. As we have done for the sine and cosine functions, we can quote the general solutions to equations involving this function.

How To: Solving Simple Equations Involving the Tangent Function

If 𝜃=𝜃 is a solution to the equation tan𝜃=𝑐, for some constant 𝑐, the set of all solutions to tan𝜃=𝑐 is 𝜃=𝜃+𝑛𝜋,𝑛.where

If 𝜃 is measured in degrees, the solution is 𝜃=𝜃+180𝑛,𝑛.where

So far, we have only considered the “standard” trigonometric functions: sine, cosine, and tangent. It is important to understand that the process holds for the reciprocal functions: cosecant, secant, and cotangent. We will demonstrate this in the next example.

Example 6: Finding the General Solution to a Reciprocal Trigonometric Equation

Find the general solution to the equation sec𝜃=2.

Answer

Recall that the secant function is the reciprocal of the cosine function. In other words, seccos𝜃1𝜃.

Hence, the equation sec𝜃=2 can be rewritten as 1𝜃=2𝜃=12=22.coscos

We know that for an angle 𝜃 measured in radians, the following exact values of the cosine function apply.

𝜃0𝜋6𝜋4𝜋3𝜋2
cos𝜃03222120

While we observe the value of cos𝜋4 to be 22, there is no value for 22. Instead, we will sketch the graph of 𝑦=𝜃cos to deduce the relevant solutions.

The graph has rotational symmetry between 0𝛼180 about 𝜋2,0, so the first solution to cos𝜃=22 is 𝜃=𝜋𝜋4=3𝜋4.

Similarly, the next solution is given by 𝜃=3𝜋2𝜋4=5𝜋4.

Since the cosine function is periodic with a period of 2𝜋 radians, further solutions are found by adding integer multiples to either of these solutions. For instance, we could find another solution by subtracting 2𝜋 from 5𝜋4: 𝜃=5𝜋42𝜋=3𝜋4.

In other words, the general solution to sec𝜃=2 is 3𝜋4+2𝜋𝑛,5𝜋4+2𝜋𝑛,𝑛.where

Alternatively, 3𝜋4+2𝜋𝑛,3𝜋4+2𝜋𝑛,𝑛.where

In this explainer, we have demonstrated how to use the symmetry and periodicity of the sine, cosine, and tangent functions to find all solutions to a trigonometric equation over a given interval or its general solution. We will now recap the key concepts.

Key Points

  • We can solve simple trigonometric equations using tables of exact values or inverse trigonometric functions.
  • To help us calculate all solutions to a given equation in a specified range, we can draw the graph of the necessary trigonometric function or use the unit circle.
  • The symmetry and periodicity of the sine, cosine, and tangent functions allow us to calculate further solutions to trigonometric equations or general solutions involving integer multiples of 360 or 2𝜋 for sine and cosine and 180 and 𝜋 for tangent.

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