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3.3: Solving Trigonometric Equations


Learning Objectives

  • Use the fundamental identities to solve trigonometric equations.
  • Express trigonometric expressions in simplest form.
  • Solve trigonometric equations by factoring.
  • Solve trigonometric equations by using the Quadratic Formula.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is (2pi). In other words, every (2pi) units, the y-values repeat. If we need to find all possible solutions, then we must add (2pi k),where (k) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is (2pi):

[sin heta=sin( heta pm 2kpi)]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example (PageIndex{1A}): Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation (cos heta=dfrac{1}{2}).

Solution

From the unit circle, we know that

[ egin{align*} cos heta &=dfrac{1}{2} [4pt] heta &=dfrac{pi}{3},space dfrac{5pi}{3} end{align*}]

These are the solutions in the interval ([ 0,2pi ]). All possible solutions are given by

[ heta=dfrac{pi}{3} pm 2kpi quad ext{and} quad heta=dfrac{5pi}{3} pm 2kpi onumber]

where (k) is an integer.

Example (PageIndex{1B}): Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation (sin t=dfrac{1}{2}).

Solution

Solving for all possible values of (t) means that solutions include angles beyond the period of (2pi). From the section on Sum and Difference Identities, we can see that the solutions are (t=dfrac{pi}{6}) and (t=dfrac{5pi}{6}). But the problem is asking for all possible values that solve the equation. Therefore, the answer is

[t=dfrac{pi}{6}pm 2pi k quad ext{and} quad t=dfrac{5pi}{6}pm 2pi k onumber]

where (k) is an integer.

How to: Given a trigonometric equation, solve using algebra

  1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
  2. Substitute the trigonometric expression with a single variable, such as (x) or (u).
  3. Solve the equation the same way an algebraic equation would be solved.
  4. Substitute the trigonometric expression back in for the variable in the resulting expressions.
  5. Solve for the angle.

Example (PageIndex{2}): Solve the Linear Trigonometric Equation

Solve the equation exactly: (2 cos heta−3=−5), (0≤ heta<2pi).

Solution

Use algebraic techniques to solve the equation.

[egin{align*} 2 cos heta-3&= -5 2 cos heta&= -2 cos heta&= -1 heta&= pi end{align*}]

Exercise (PageIndex{1})

Solve exactly the following linear equation on the interval ([0,2pi)): (2 sin x+1=0).

Answer

(x=dfrac{7pi}{6},space dfrac{11pi}{6})

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is (pi),not (2pi). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of (dfrac{pi}{2}),unless, of course, a problem places its own restrictions on the domain.

Example (PageIndex{3A}): Solving a Trignometric Equation Involving Sine

Solve the problem exactly: (2 {sin}^2 heta−1=0), (0≤ heta<2pi).

Solution

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate (sin heta). Then we will find the angles.

[egin{align*}
2 {sin}^2 heta-1&= 0
2 {sin}^2 heta&= 1
{sin}^2 heta&= dfrac{1}{2}
sqrt{ {sin}^2 heta }&= pm sqrt{ dfrac{1}{2} }
sin heta&= pm dfrac{1}{sqrt{2}}
&= pm dfrac{sqrt{2}}{2}
heta&= dfrac{pi}{4}, space dfrac{3pi}{4},space dfrac{5pi}{4}, space dfrac{7pi}{4}
end{align*}]

Analysis

As (sin heta=−dfrac{1}{2}), notice that all four solutions are in the third and fourth quadrants.

Example (PageIndex{3B}): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: (csc heta=−2), (0≤ heta<4pi).

Solution

We want all values of ( heta) for which (csc heta=−2) over the interval (0≤ heta<4pi).

[egin{align*} csc heta&= -2 dfrac{1}{sin heta}&= -2 sin heta&= -dfrac{1}{2} heta&= dfrac{7pi}{6},space dfrac{11pi}{6},space dfrac{19pi}{6}, space dfrac{23pi}{6} end{align*}]

Example (PageIndex{3C}): Solving an Equation Involving Tangent

Solve the equation exactly: ( anleft( heta−dfrac{pi}{2} ight)=1), (0≤ heta<2pi).

Solution

Recall that the tangent function has a period of (pi). On the interval ([ 0,pi )),and at the angle of (dfrac{pi}{4}),the tangent has a value of (1). However, the angle we want is (left( heta−dfrac{pi}{2} ight)). Thus, if ( anleft(dfrac{pi}{4} ight)=1),then

[egin{align*} heta-dfrac{pi}{2}&= dfrac{pi}{4} heta&= dfrac{3pi}{4} pm kpi end{align*}]

Over the interval ([ 0,2pi )),we have two solutions:

( heta=dfrac{3pi}{4}) and ( heta=dfrac{3pi}{4}+pi=dfrac{7pi}{4})

Exercise (PageIndex{2})

Find all solutions for ( an x=sqrt{3}).

Answer

(dfrac{pi}{3}pm pi k)

Example (PageIndex{4}): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation (2( an x+3)=5+ an x), (0≤x<2pi).

Solution

We can solve this equation using only algebra. Isolate the expression ( an x) on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of (−1): ( heta=dfrac{3pi}{4}) and ( heta=dfrac{7pi}{4}).

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example (PageIndex{5A}): Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation (sin heta=0.8),where ( heta) is in radians.

Solution

Make sure mode is set to radians. To find ( heta), use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the ({sin}^{−1}) function. What is shown on the screen is ({sin}^{−1}).The calculator is ready for the input within the parentheses. For this problem, we enter ({sin}^{−1}(0.8)), and press ENTER. Thus, to four decimals places,

({sin}^{−1}(0.8)≈0.9273)

The solution is

( heta≈0.9273pm 2pi k)

The angle measurement in degrees is

[egin{align*} heta&approx 53.1^{circ} heta&approx 180^{circ}-53.1^{circ} &approx 126.9^{circ} end{align*}]

Analysis

Note that a calculator will only return an angle in quadrants I or IV for the sine function since that is the range of the inverse sine. The other angle is obtained by using (pi− heta).

Example (PageIndex{5B}): Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation ( sec θ=−4, ) giving your answer in radians.

Solution

We can begin with some algebra.

[egin{align*} sec heta&= -4 dfrac{1}{cos heta}&= -4 cos heta&= -dfrac{1}{4} end{align*}]

Check that the MODE is in radians. Now use the inverse cosine function

[egin{align*}{cos}^{-1}left(-dfrac{1}{4} ight)&approx 1.8235 heta&approx 1.8235+2pi k end{align*}]

Since (dfrac{pi}{2}≈1.57) and (pi≈3.14),(1.8235) is between these two numbers, thus ( heta≈1.8235) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure (PageIndex{2}).

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is ( heta '≈pi−1.8235≈1.3181). The other solution in quadrant III is ( heta '≈pi+1.3181≈4.4597).

The solutions are ( heta≈1.8235pm 2pi k) and ( heta≈4.4597pm 2pi k).

Exercise (PageIndex{3})

Solve (cos heta=−0.2).

Answer

( heta≈1.7722pm 2pi k) and ( heta≈4.5110pm 2pi k)

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as (x) or (u). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example (PageIndex{6A}): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: ({cos}^2 heta+3 cos heta−1=0), (0≤ heta<2pi).

Solution

We begin by using substitution and replacing (cos heta) with (x). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let (cos heta=x). We have

(x^2+3x−1=0)

The equation cannot be factored, so we will use the quadratic formula: (x=dfrac{−bpm sqrt{b^2−4ac}}{2a}).

[egin{align*} x&= dfrac{ -3pm sqrt{ {(-3)}^2-4 (1) (-1) } }{2} &= dfrac{-3pm sqrt{13}}{2}end{align*}]

Replace (x) with (cos heta ) and solve.

[egin{align*} cos heta&= dfrac{-3pm sqrt{13}}{2} heta&= {cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) end{align*}]

Note that only the + sign is used. This is because we get an error when we solve ( heta={cos}^{−1}left(dfrac{−3−sqrt{13}}{2} ight)) on a calculator, since the domain of the inverse cosine function is ([ −1,1 ]). However, there is a second solution:

[egin{align*} heta&= {cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) &approx 1.26 end{align*}]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

[egin{align*} heta&= 2pi-{cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) &approx 5.02 end{align*}]

Example (PageIndex{6B}): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: (2 {sin}^2 heta−5 sin heta+3=0), (0≤ heta≤2pi).

Solution

Using grouping, this quadratic can be factored. Either make the real substitution, (sin heta=u),or imagine it, as we factor:

[egin{align*} 2 {sin}^2 heta-5 sin heta+3&= 0 (2 sin heta-3)(sin heta-1)&= 0 qquad ext {Now set each factor equal to zero.} 2 sin heta-3&= 0 2 sin heta&= 3 sin heta&= dfrac{3}{2} sin heta-1&= 0 sin heta&= 1 end{align*}]

Next solve for ( heta): (sin heta≠dfrac{3}{2}), as the range of the sine function is ([ −1,1 ]). However, (sin heta=1), giving the solution ( heta=dfrac{pi}{2}).

Analysis

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise (PageIndex{4})

Solve ({sin}^2 heta=2 cos heta+2), (0≤ heta≤2pi). [Hint: Make a substitution to express the equation only in terms of cosine.]

Answer

(cos heta=−1), ( heta=pi)

Example (PageIndex{7A}): Solving a Trigonometric Equation Using Algebra

Solve exactly: (2 {sin}^2 heta+sin heta=0;space 0≤ heta<2pi)

Solution

This problem should appear familiar as it is similar to a quadratic. Let (sin heta=x). The equation becomes (2x^2+x=0). We begin by factoring:

[egin{align*}
2x^2+x&= 0
x(2x+1)&= 0qquad ext {Set each factor equal to zero.}
x&= 0
2x+1&= 0
x&= -dfrac{1}{2} end{align*}]
Then, substitute back into the equation the original expression (sin heta ) for (x). Thus,
[egin{align*} sin heta&= 0
heta&= 0,pi
sin heta&= -dfrac{1}{2}
heta&= dfrac{7pi}{6},dfrac{11pi}{6}
end{align*}]

The solutions within the domain (0≤ heta<2pi) are ( heta=0,pi,dfrac{7pi}{6},dfrac{11pi}{6}).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[egin{align*} {sin}^2 heta+sin heta&= 0 sin heta(2sin heta+1)&= 0 sin heta&= 0 heta&= 0,pi 2 sin heta+1&= 0 2sin heta&= -1 sin heta&= -dfrac{1}{2} heta&= dfrac{7pi}{6},dfrac{11pi}{6} end{align*}]

Analysis

We can see the solutions on the graph in Figure (PageIndex{3}). On the interval (0≤ heta<2pi),the graph crosses the (x)-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example (PageIndex{7B}): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: (2 {sin}^2 heta−3 sin heta+1=0), (0≤ heta<2pi).

Solution

We can factor using grouping. Solution values of ( heta) can be found on the unit circle.

[egin{align*} (2 sin heta-1)(sin heta-1)&= 0 2 sin heta-1&= 0 sin heta&= dfrac{1}{2} heta&= dfrac{pi}{6}, dfrac{5pi}{6} sin heta&= 1 heta&= dfrac{pi}{2} end{align*}]

Exercise (PageIndex{5})

Solve the quadratic equation (2{cos}^2 heta+cos heta=0).

Answer

(dfrac{pi}{2}, space dfrac{2pi}{3}, space dfrac{4pi}{3}, space dfrac{3pi}{2})

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example (PageIndex{8}): Solving an Equation Using an Identity

Solve the equation exactly using an identity: (3 cos heta+3=2 {sin}^2 heta), (0≤ heta<2pi).

Solution

If we rewrite the right side, we can write the equation in terms of cosine:

[egin{align*}
3 cos heta+3&= 2 {sin}^2 heta
3 cos heta+3&= 2(1-{cos}^2 heta)
3 cos heta+3&= 2-2{cos}^2 heta
2 {cos}^2 heta+3 cos heta+1&= 0
(2 cos heta+1)(cos heta+1)&= 0
2 cos heta+1&= 0
cos heta&= -dfrac{1}{2}
heta&= dfrac{2pi}{3},space dfrac{4pi}{3}
cos heta+1&= 0
cos heta&= -1
heta&= pi
end{align*}]

Our solutions are ( heta=dfrac{2pi}{3},space dfrac{4pi}{3},space pi).

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as (sin(2x)) or (cos(3x)). When confronted with these equations, recall that (y=sin(2x)) is a horizontal compression by a factor of 2 of the function (y=sin x). On an interval of (2pi),we can graph two periods of (y=sin(2x)),as opposed to one cycle of (y=sin x). This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to (sin(2x)=0) compared to (sin x=0). This information will help us solve the equation.

Example (PageIndex{9}): Solving a Multiple Angle Trigonometric Equation

Solve exactly: (cos(2x)=dfrac{1}{2}) on ([ 0,2pi )).

Solution

We can see that this equation is the standard equation with a multiple of an angle. If (cos(alpha)=dfrac{1}{2}),we know (alpha) is in quadrants I and IV. While ( heta={cos}^{−1} dfrac{1}{2}) will only yield solutions in quadrants I and II, we recognize that the solutions to the equation (cos heta=dfrac{1}{2}) will be in quadrants I and IV.

Therefore, the possible angles are ( heta=dfrac{pi}{3}) and ( heta=dfrac{5pi}{3}). So, (2x=dfrac{pi}{3}) or (2x=dfrac{5pi}{3}), which means that (x=dfrac{pi}{6}) or (x=dfrac{5pi}{6}). Does this make sense? Yes, because (cosleft(2left(dfrac{pi}{6} ight) ight)=cosleft(dfrac{pi}{3} ight)=dfrac{1}{2}).

Are there any other possible answers? Let us return to our first step.

In quadrant I, (2x=dfrac{pi}{3}), so (x=dfrac{pi}{6}) as noted. Let us revolve around the circle again:

[egin{align*}
2x&= dfrac{pi}{3}+2pi
&= dfrac{pi}{3}+dfrac{6pi}{3}
&= dfrac{7pi}{3}
x&= dfrac{7pi}{6}
ext {One more rotation yields}
2x&= dfrac{pi}{3}+4pi
&= dfrac{pi}{3}+dfrac{12pi}{3}
&= dfrac{13pi}{3}
end{align*}]

(x=dfrac{13pi}{6}>2pi), so this value for (x) is larger than (2pi), so it is not a solution on ([ 0,2pi )).

In quadrant IV, (2x=dfrac{5pi}{3}), so (x=dfrac{5pi}{6}) as noted. Let us revolve around the circle again:

[egin{align*} 2x&= dfrac{5pi}{3}+2pi &= dfrac{5pi}{3}+dfrac{6pi}{3} &= dfrac{11pi}{3} end{align*}]

so (x=dfrac{11pi}{6}).

One more rotation yields

[egin{align*} 2x&= dfrac{5pi}{3}+4pi &= dfrac{5pi}{3}+dfrac{12pi}{3} &= dfrac{17pi}{3} end{align*}]

(x=dfrac{17pi}{6}>2pi),so this value for (x) is larger than (2pi),so it is not a solution on ([ 0,2pi )).

Our solutions are (x=dfrac{pi}{6}, space dfrac{5pi}{6}, space dfrac{7pi}{6}), and (dfrac{11pi}{6}). Note that whenever we solve a problem in the form of (sin(nx)=c), we must go around the unit circle (n) times.

Key Concepts

  • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example (PageIndex{1}), Example (PageIndex{2}), and Example (PageIndex{3}).
  • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example (PageIndex{4}), Example (PageIndex{5}), and Example (PageIndex{6}), and Example (PageIndex{7}).
  • We can also solve trigonometric equations using a graphing calculator. See Example (PageIndex{8}) and Example (PageIndex{9}).
  • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example (PageIndex{10}), Example (PageIndex{11}), Example (PageIndex{12}), and Example (PageIndex{13}).
  • We can also use the identities to solve trigonometric equation. See Example (PageIndex{14}), Example (PageIndex{15}), and Example (PageIndex{16}).
  • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example (PageIndex{17}).
  • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example (PageIndex{18}).


Watch the video: L3 How to solve trigonometric equations general solutions (November 2021).