9.6: Solving Trigonometric Equations (2024)

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    Learning Objectives

    In this section, you will:

    • Solve linear trigonometric equations in sine and cosine.
    • Solve equations involving a single trigonometric function.
    • Solve trigonometric equations using a calculator.
    • Solve trigonometric equations that are quadratic in form.
    • Solve trigonometric equations using fundamentalidentities.
    • Solve trigonometric equations with multiple angles.
    • Solve right triangle problems.
    9.6: Solving Trigonometric Equations (2)

    Figure1Egyptianpyramids standing near a modern city. (credit: Oisin Mulvihill)

    Thales of Miletus (circa 625–547 BC) is known as the founder ofgeometry. The legend is that he calculated the height of the GreatPyramid of Giza in Egypt using the theory ofsimilar triangles, which he developed by measuringthe shadow of his staff. He reasoned that when the height of hisstaff's shadow was exactly equal to the actual height of the staff,then the height of the nearby pyramid's shadow must also be equalto the height of the actual pyramid. Since the structures and theirshadows were creating a right triangle with two equal sides, theywere similar triangles. By measuring the length of the pyramid'sshadow at that moment, he could obtain the height of the pyramid.Based on proportions, this theory has applications in a number ofareas, including fractal geometry, engineering, and architecture.Often, the angle of elevation and the angle of depression are foundusing similar triangles.

    In earlier sections of this chapter, we looked at trigonometricidentities. Identities are true for all values in the domain of thevariable. In this section, we begin our study of trigonometricequations to study real-world scenarios such as the finding thedimensions of the pyramids.

    Solving Linear TrigonometricEquations in Sine and Cosine

    Trigonometric equations are, as the name implies, equations thatinvolve trigonometric functions. Similar in many ways to solvingpolynomial equations or rational equations, only specific values ofthe variable will be solutions, if there are solutions at all.Often we will solve a trigonometric equation over a specifiedinterval. However, just as often, we will be asked to find allpossible 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 functionmust be considered before we assume that any solution is valid.Theperiodof both the sine function and the cosinefunction is2π.2π.In other words,every2π2πunits, they-values repeat. If we need to find all possiblesolutions, then we mustadd2πk,2πk,wherekkis an integer, to theinitial solution. Recall the rule that gives the format for statingall possible solutions for a function where the periodis2π:2π:

    sinθ=sin(θ±2kπ)sinθ=sin(θ±2kπ)

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

    EXAMPLE1

    Solving a Linear TrigonometricEquation Involving the Cosine Function

    Find all possible exact solutions for theequationcosθ=12.cosθ=12.

    Answer

    EXAMPLE2

    Solving a Linear Equation Involvingthe Sine Function

    Find all possible exact solutions for theequationsint=12.sint=12.

    Answer
    HOWTO

    Given a trigonometric equation, solve usingalgebra.

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

    EXAMPLE3

    Solve the Linear TrigonometricEquation

    Solve the equationexactly:2cosθ−3=−5,0≤θ<2π.2cosθ−3=−5,0≤θ<2π.

    Answer
    TRYIT#1

    Solve exactly the following linear equation on theinterval[0,2π):2sinx+1=0.[0,2π):2sinx+1=0.

    Solving Equations Involving a SingleTrigonometric Function

    When we are given equations that involve only one of the sixtrigonometric functions, their solutions involve using algebraictechniques and the unit circle (seeFigure2). We need to make several considerations when theequation involves trigonometric functions other than sine andcosine. Problems involving the reciprocals of the primarytrigonometric functions need to be viewed from an algebraicperspective. In other words, we will write the reciprocal function,and solve for the angles using the function. Also, an equationinvolving the tangent function is slightly different from onecontaining a sine or cosine function. First, as we know, the periodof tangent isπ,π,not2π.2π.Further, thedomain of tangent is all real numbers with the exception of oddinteger multiples ofπ2,π2,unless, of course, a problemplaces its own restrictions on the domain.

    EXAMPLE4

    Solving a Problem Involving a SingleTrigonometric Function

    Solve the problemexactly:2sin2θ−1=0,0≤θ<2π.2sin2θ−1=0,0≤θ<2π.

    Answer

    EXAMPLE5

    Solving a Trigonometric EquationInvolving Cosecant

    Solve the following equationexactly:cscθ=−2,0≤θ<4π.cscθ=−2,0≤θ<4π.

    Answer

    Analysis

    Assinθ=−12,sinθ=−12,notice that all four solutionsare in the third and fourth quadrants.

    EXAMPLE6

    Solving an Equation InvolvingTangent

    Solve the equationexactly:tan(θ−π2)=1,0≤θ<2π.tan(θ−π2)=1,0≤θ<2π.

    Answer
    TRYIT#2

    Find all solutions fortanx=3–√.tanx=3.

    EXAMPLE7

    Identify all Solutions to theEquation Involving Tangent

    Identify all exact solutions to theequation2(tanx+3)=5+tanx,0≤x<2π.2(tanx+3)=5+tanx,0≤x<2π.

    Answer

    Solve Trigonometric Equations Using aCalculator

    Not all functions can be solved exactly using only the unitcircle. When we must solve an equation involving an angle otherthan 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.

    EXAMPLE8

    Using a Calculator to Solve aTrigonometric Equation Involving Sine

    Use a calculator to solve theequationsinθ=0.8,sinθ=0.8,whereθθis inradians.

    Answer

    Analysis

    Note that a calculator will only return an angle in quadrants Ior IV for the sine function, since that is the range of the inversesine. The other angle is obtained by usingπ−θ.π−θ.Thus,the additional solution is≈2.2143±2πk≈2.2143±2πk

    EXAMPLE9

    Using a Calculator to Solve aTrigonometric Equation Involving Secant

    Use a calculator to solve theequationsecθ=−4,secθ=−4,giving your answer inradians.

    Answer
    9.6: Solving Trigonometric Equations (3)
    TRYIT#3

    Solvecosθ=−0.2.cosθ=−0.2.

    Solving Trigonometric Equations inQuadratic Form

    Solving aquadratic equationmay be more complicated,but once again, we can use algebra as we would for any quadraticequation. Look at the pattern of the equation. Is there more thanone trigonometric function in the equation, or is there only one?Which trigonometric function is squared? If there is only onefunction represented and one of the terms is squared, think aboutthe standard form of a quadratic. Replace the trigonometricfunction with a variable such asxxoru.u.Ifsubstitution makes the equation look like a quadratic equation,then we can use the same methods for solving quadratics to solvethe trigonometric equations.

    EXAMPLE10

    Solving a Trigonometric Equation inQuadratic Form

    Solve the equationexactly:cos2θ+3cosθ−1=0,0≤θ<2π.cos2θ+3cosθ−1=0,0≤θ<2π.

    Answer

    EXAMPLE11

    Solving a Trigonometric Equation inQuadratic Form by Factoring

    Solve the equationexactly:2sin2θ−5sinθ+3=0,0≤θ≤2π.2sin2θ−5sinθ+3=0,0≤θ≤2π.

    Answer

    Analysis

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

    TRYIT#4

    Solvesin2θ=2cosθ+2,0≤θ≤2π.sin2θ=2cosθ+2,0≤θ≤2π.[Hint:Make a substitution to express the equation only in terms ofcosine.]

    EXAMPLE12

    Solving a Trigonometric EquationUsing Algebra

    Solve exactly:

    2sin2θ+sinθ=0;0≤θ<2π2sin2θ+sinθ=0;0≤θ<2π

    Answer

    Analysis

    We can see the solutions on the graph inFigure3. On the interval0≤θ<2π,0≤θ<2π,the graphcrosses thex-axis four times,at the solutions noted. Notice that trigonometric equations thatare in quadratic form can yield up to four solutions instead of theexpected two that are found with quadratic equations. In thisexample, each solution (angle) corresponding to a positive sinevalue will yield two angles that would result in that value.

    9.6: Solving Trigonometric Equations (4)

    Figure3

    We can verify the solutions on the unit circleinFigure2as well.

    EXAMPLE13

    Solving a Trigonometric EquationQuadratic in Form

    Solve the equation quadratic in formexactly:2sin2θ−3sinθ+1=0,0≤θ<2π.2sin2θ−3sinθ+1=0,0≤θ<2π.

    Answer
    TRYIT#5

    Solve the quadraticequation2cos2θ+cosθ=0.2cos2θ+cosθ=0.

    Solving Trigonometric Equations UsingFundamental Identities

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

    EXAMPLE14

    Use Identities to Solve anEquation

    Use identities to solve exactly the trigonometric equation overthe interval0≤x<2π.0≤x<2π.

    cosxcos(2x)+sinxsin(2x)=3–√2cosxcos(2x)+sinxsin(2x)=32

    Answer

    EXAMPLE15

    Solving the Equation Using aDouble-Angle Formula

    Solve the equation exactly using a double-angleformula:cos(2θ)=cosθ.cos(2θ)=cosθ.

    Answer

    EXAMPLE16

    Solving an Equation Using anIdentity

    Solve the equation exactly using anidentity:3cosθ+3=2sin2θ,0≤θ<2π.3cosθ+3=2sin2θ,0≤θ<2π.

    Answer

    Solving Trigonometric Equations withMultiple Angles

    Sometimes it is not possible to solve a trigonometric equationwith identities that have a multiple angle, suchassin(2x)sin(2x)orcos(3x).cos(3x).Whenconfronted with these equations, recallthaty=sin(2x)y=sin(2x)is ahorizontalcompressionby a factor of 2 of thefunctiony=sinx.y=sinx.On an intervalof2π,2π,we can graph two periodsofy=sin(2x),y=sin(2x),as opposed to one cycleofy=sinx.y=sinx.This compression of the graph leads usto believe there may be twice as manyx-intercepts or solutionstosin(2x)=0sin(2x)=0comparedtosinx=0.sinx=0.This information will help us solve theequation.

    EXAMPLE17

    Solving a Multiple AngleTrigonometric Equation

    Solveexactly:cos(2x)=12cos(2x)=12on[0,2π).[0,2π).

    Answer

    Solving Right TriangleProblems

    We can now use all of the methods we have learned to solveproblems that involve applying the properties of right trianglesand thePythagorean Theorem. We begin with the familiarPythagorean Theorem,a2+b2=c2,a2+b2=c2,and model anequation to fit a situation.

    EXAMPLE18

    Using the Pythagorean Theorem toModel an Equation

    Use the Pythagorean Theorem, and the properties of righttriangles to model an equation that fits the problem.

    One of the cables that anchors the center of the London EyeFerris wheel to the ground must be replaced. The center of theFerris wheel is 69.5 meters above the ground, and the second anchoron the ground is 23 meters from the base of the Ferris wheel.Approximately how long is the cable, and what is the angle ofelevation (from ground up to the center of the Ferris wheel)?SeeFigure4.

    9.6: Solving Trigonometric Equations (5)

    Figure4

    Answer

    EXAMPLE19

    Using the Pythagorean Theorem toModel an Abstract Problem

    OSHA safety regulations require that the base of a ladder beplaced 1 foot from the wall for every 4 feet of ladder length. Findthe angle that a ladder of any length forms with the ground and theheight at which the ladder touches the wall.

    Answer
    9.6: Solving Trigonometric Equations (6)

    9.5 SectionExercises

    Verbal

    1.

    Will there always be solutions to trigonometric functionequations? If not, describe an equation that would not have asolution. Explain why or why not.

    2.

    When solving a trigonometric equation involving more than onetrig function, do we always want to try to rewrite the equation soit is expressed in terms of one trigonometric function? Why or whynot?

    3.

    When solving linear trig equations in terms of only sine orcosine, how do we know whether there will be solutions?

    Algebraic

    For the following exercises, find all solutions exactly on theinterval0≤θ<2π.0≤θ<2π.

    4.

    2sinθ=−2–√2sinθ=−2

    5.

    2sinθ=3–√2sinθ=3

    6.

    2cosθ=12cosθ=1

    7.

    2cosθ=−2–√2cosθ=−2

    8.

    tanθ=−1tanθ=−1

    9.

    tanx=1tanx=1

    10.

    cotx+1=0cotx+1=0

    11.

    4sin2x−2=04sin2x−2=0

    12.

    csc2x−4=0csc2x−4=0

    For the following exercises, solve exactlyon[0,2π).[0,2π).

    13.

    2cosθ=2–√2cosθ=2

    14.

    2cosθ=−12cosθ=−1

    15.

    2sinθ=−12sinθ=−1

    16.

    2sinθ=−3–√2sinθ=−3

    17.

    2sin(3θ)=12sin(3θ)=1

    18.

    2sin(2θ)=3–√2sin(2θ)=3

    19.

    2cos(3θ)=−2–√2cos(3θ)=−2

    20.

    cos(2θ)=−3√2cos(2θ)=−32

    21.

    2sin(πθ)=12sin(πθ)=1

    22.

    2cos(π5θ)=3–√2cos(π5θ)=3

    For the following exercises, find all exact solutionson[0,2π).[0,2π).

    23.

    sec(x)sin(x)−2sin(x)=0sec(x)sin(x)−2sin(x)=0

    24.

    tan(x)−2sin(x)tan(x)=0tan(x)−2sin(x)tan(x)=0

    25.

    2cos2t+cos(t)=12cos2t+cos(t)=1

    26.

    2tan2(t)=3sec(t)2tan2(t)=3sec(t)

    27.

    2sin(x)cos(x)−sin(x)+2cos(x)−1=02sin(x)cos(x)−sin(x)+2cos(x)−1=0

    28.

    cos2θ=12cos2θ=12

    29.

    sec2x=1sec2x=1

    30.

    tan2(x)=−1+2tan(−x)tan2(x)=−1+2tan(−x)

    31.

    8sin2(x)+6sin(x)+1=08sin2(x)+6sin(x)+1=0

    32.

    tan5(x)=tan(x)tan5(x)=tan(x)

    For the following exercises, solve with the methods shown inthis section exactly on the interval[0,2π).[0,2π).

    33.

    sin(3x)cos(6x)−cos(3x)sin(6x)=−0.9sin(3x)cos(6x)−cos(3x)sin(6x)=−0.9

    34.

    sin(6x)cos(11x)−cos(6x)sin(11x)=−0.1sin(6x)cos(11x)−cos(6x)sin(11x)=−0.1

    35.

    cos(2x)cosx+sin(2x)sinx=1cos(2x)cosx+sin(2x)sinx=1

    36.

    6sin(2t)+9sint=06sin(2t)+9sint=0

    37.

    9cos(2θ)=9cos2θ−49cos(2θ)=9cos2θ−4

    38.

    sin(2t)=costsin(2t)=cost

    39.

    cos(2t)=sintcos(2t)=sint

    40.

    cos(6x)−cos(3x)=0cos(6x)−cos(3x)=0

    For the following exercises, solve exactly on theinterval[0,2π).[0,2π).Use the quadraticformula if the equations do not factor.

    41.

    tan2x−3–√tanx=0tan2x−3tanx=0

    42.

    sin2x+sinx−2=0sin2x+sinx−2=0

    43.

    sin2x−2sinx−4=0sin2x−2sinx−4=0

    44.

    5cos2x+3cosx−1=05cos2x+3cosx−1=0

    45.

    3cos2x−2cosx−2=03cos2x−2cosx−2=0

    46.

    5sin2x+2sinx−1=05sin2x+2sinx−1=0

    47.

    tan2x+5tanx−1=0tan2x+5tanx−1=0

    48.

    cot2x=−cotxcot2x=−cotx

    49.

    −tan2x−tanx−2=0−tan2x−tanx−2=0

    For the following exercises, find exact solutions on theinterval[0,2π).[0,2π).Look for opportunities to usetrigonometric identities.

    50.

    sin2x−cos2x−sinx=0sin2x−cos2x−sinx=0

    51.

    sin2x+cos2x=0sin2x+cos2x=0

    52.

    sin(2x)−sinx=0sin(2x)−sinx=0

    53.

    cos(2x)−cosx=0cos(2x)−cosx=0

    54.

    2tanx2−sec2x−sin2x=cos2x2tanx2−sec2x−sin2x=cos2x

    55.

    1−cos(2x)=1+cos(2x)1−cos(2x)=1+cos(2x)

    56.

    sec2x=7sec2x=7

    57.

    10sinxcosx=6cosx10sinxcosx=6cosx

    58.

    −3sint=15costsint−3sint=15costsint

    59.

    4cos2x−4=15cosx4cos2x−4=15cosx

    60.

    8sin2x+6sinx+1=08sin2x+6sinx+1=0

    61.

    8cos2θ=3−2cosθ8cos2θ=3−2cosθ

    62.

    6cos2x+7sinx−8=06cos2x+7sinx−8=0

    63.

    12sin2t+cost−6=012sin2t+cost−6=0

    64.

    tanx=3sinxtanx=3sinx

    65.

    cos3t=costcos3t=cost

    Graphical

    For the following exercises, algebraically determine allsolutions of the trigonometric equation exactly, then verify theresults by graphing the equation and finding the zeros.

    66.

    6sin2x−5sinx+1=06sin2x−5sinx+1=0

    67.

    8cos2x−2cosx−1=08cos2x−2cosx−1=0

    68.

    100tan2x+20tanx−3=0100tan2x+20tanx−3=0

    69.

    2cos2x−cosx+15=02cos2x−cosx+15=0

    70.

    20sin2x−27sinx+7=020sin2x−27sinx+7=0

    71.

    2tan2x+7tanx+6=02tan2x+7tanx+6=0

    72.

    130tan2x+69tanx−130=0130tan2x+69tanx−130=0

    Technology

    For the following exercises, use a calculator to find allsolutions to four decimal places.

    73.

    sinx=0.27sinx=0.27

    74.

    sinx=−0.55sinx=−0.55

    75.

    tanx=−0.34tanx=−0.34

    76.

    cosx=0.71cosx=0.71

    For the following exercises, solve the equations algebraically,and then use a calculator to find the values on theinterval[0,2π).[0,2π).Round to four decimal places.

    77.

    tan2x+3tanx−3=0tan2x+3tanx−3=0

    78.

    6tan2x+13tanx=−66tan2x+13tanx=−6

    79.

    tan2x−secx=1tan2x−secx=1

    80.

    sin2x−2cos2x=0sin2x−2cos2x=0

    81.

    2tan2x+9tanx−6=02tan2x+9tanx−6=0

    82.

    4sin2x+sin(2x)secx−3=04sin2x+sin(2x)secx−3=0

    Extensions

    For the following exercises, find all solutions exactly to theequations on the interval[0,2π).[0,2π).

    83.

    csc2x−3cscx−4=0csc2x−3cscx−4=0

    84.

    sin2x−cos2x−1=0sin2x−cos2x−1=0

    85.

    sin2x(1−sin2x)+cos2x(1−sin2x)=0sin2x(1−sin2x)+cos2x(1−sin2x)=0

    86.

    3sec2x+2+sin2x−tan2x+cos2x=03sec2x+2+sin2x−tan2x+cos2x=0

    87.

    sin2x−1+2cos(2x)−cos2x=1sin2x−1+2cos(2x)−cos2x=1

    88.

    tan2x−1−sec3xcosx=0tan2x−1−sec3xcosx=0

    89.

    sin(2x)sec2x=0sin(2x)sec2x=0

    90.

    sin(2x)2csc2x=0sin(2x)2csc2x=0

    91.

    2cos2x−sin2x−cosx−5=02cos2x−sin2x−cosx−5=0

    92.

    1sec2x+2+sin2x+4cos2x=41sec2x+2+sin2x+4cos2x=4

    Real-World Applications

    93.

    An airplane has only enough gas to fly to a city 200 milesnortheast of its current location. If the pilot knows that the cityis 25 miles north, how many degrees north of east should theairplane fly?

    94.

    If a loading ramp is placed next to a truck, at a height of 4feet, and the ramp is 15 feet long, what angle does the ramp makewith the ground?

    95.

    If a loading ramp is placed next to a truck, at a height of 2feet, and the ramp is 20 feet long, what angle does the ramp makewith the ground?

    96.

    A woman is watching a launched rocket currently 11 miles inaltitude. If she is standing 4 miles from the launch pad, at whatangle is she looking up from horizontal?

    97.

    An astronaut is in a launched rocket currently 15 miles inaltitude. If a man is standing 2 miles from the launch pad, at whatangle is the astronaut looking down at him from horizontal? (Hint:this is called the angle of depression.)

    98.

    A woman is standing 8 meters away from a 10-meter tall building.At what angle is she looking to the top of the building?

    99.

    Issa is standing 10 meters away from a 6-meter tall building.Travis is at the top of the building looking down at Issa. At whatangle is Travis looking at Issa?

    100.

    A 20-foot tall building has a shadow that is 55 feet long. Whatis the angle of elevation of the sun?

    101.

    A 90-foot tall building has a shadow that is 2 feet long. Whatis the angle of elevation of the sun?

    102.

    A spotlight on the ground 3 meters from a 2-meter tall man castsa 6 meter shadow on a wall 6 meters from the man. At what angle isthe light?

    103.

    A spotlight on the ground 3 feet from a 5-foot tall woman castsa 15-foot tall shadow on a wall 6 feet from the woman. At whatangle is the light?

    For the following exercises, find a solution to the followingword problem algebraically. Then use a calculator to verify theresult. Round the answer to the nearest tenth of a degree.

    104.

    A person does a handstand with their feet touching a wall andtheir hands 1.5 feet away from the wall. If the person is 6 feettall, what angle do their feet make with the wall?

    105.

    A person does a handstand with her feet touching a wall and herhands 3 feet away from the wall. If the person is 5 feet tall, whatangle do her feet make with the wall?

    106.

    A 23-foot ladder is positioned next to a house. If the ladderslips at 7 feet from the house when there is not enough traction,what angle should the ladder make with the ground to avoidslipping?

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    9.6: Solving Trigonometric Equations (2024)
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