trigonometry                                                                           trigonometry
                                                                                                                   Solving Trigonometric Equations
                            MHF4U: Advanced Functions                                                              Like other types of equations (linear, quadratic, etc.), we can
                                                                                                                   solve trigonometric equations by isolating the independent
                                                                                                                   variable.
                                                                                                                   In some cases, this can be done using simple algebra,
                                                                                                                   whereas in other cases, it may be necessary to factor a
                            Solving Trigonometric Equations                                                        trigonometric expression first.
                            J. Garvin                                                                              When possible, use exact values to preserve accuracy.
                                                                                                                   J. Garvin — Solving Trigonometric Equations
                            Slide 1/12                                                                             Slide 2/12
                                                                                        trigonometry                                                                           trigonometry
                            Solving Trigonometric Equations                                                        Solving Trigonometric Equations
                            Example                                                                                Example
                            Solve 2sinx −1 = 0 on the interval [0,2π].                                             Solve 3tanx +2 = 0 on the interval [0,2π].
                                                   2sinx = 1                                                                             3tanx = −2
                                                     sinx = 1                                                                             tanx = −2
                                                              2     
                                                                                3       

                                                                −1 1                                                                                   −1    2
                                                        x = sin      2                                                                        x = tan      −3
                            Since sine is positive in quadrants 1 and 2, there are two                             Tangent is negative in quadrants 2 and 4.
                            solutions.                                                                             Calculating tan−1−2
 yields approximately −0.588, which
                                                  π                                  π     5π                                             3
                            In quadrant 1, x = 6, while in quadrant 2, x = π − 6 = 6 .                             is a negative rotation into quadrant 4.
                                                                                                                   In quadrant 2, x ≈ π −0.588 ≈ 2.55, while in quadrant 4,
                                                                                                                   x ≈ 2π −0.588 ≈ 5.70.
                            J. Garvin — Solving Trigonometric Equations                                            J. Garvin — Solving Trigonometric Equations
                            Slide 3/12                                                                             Slide 4/12
                                                                                        trigonometry                                                                           trigonometry
                            Solving Trigonometric Equations                                                        Solving Trigonometric Equations
                            Example                                                                                Example
                                       2                                                                                       2
                            Solve 4sin x −3 = 0 on the interval [0,2π].                                            Solve 5cos x −17cosx +6 = 0 on the interval [0,2π].
                            Factor as a difference of squares.                                                      Factor as a complex trinomial.
                                       √               √
                             (2sinx −     3)(2sinx +     3) = 0      √               √                               (5cosx −2)(cosx −3) = 0              

                                                                  −1    3          −1       3                                                    x = cos−1 2      or cos−1(3)
                                                          x = sin      2    or sin      − 2                                                                   5
                                                       √                                                                                 −12

                                                    −1    3                                                        In the case of x = cos       5 , cosine is positive in quadrants
                            In the case of x = sin       2   , sine is positive in quadrants 1
                            and 2, so x = π or x = 2π.                                                             1 and 4, so x ≈ 1.16 or x ≈ 2π − 1.16 ≈ 5.12.
                                            3          3                                                                                 −1
                                                        √                                                        The case of x = cos      (3) is not possible, since cosine has a
                                                    −1      3
                            In the case of x = sin      − 2 , sine is negative in                                  range of [−1,1].
                            quadrants 3 and 4, so x = 4π or x = 5π.                                                Thus, only two solutions are admissible.
                                                          3           3
                            All four solutions are possible.
                            J. Garvin — Solving Trigonometric Equations                                            J. Garvin — Solving Trigonometric Equations
                            Slide 5/12                                                                             Slide 6/12
                                                                                       trigonometry                                                                          trigonometry
                            Solving Trigonometric Equations                                                       Solving Trigonometric Equations
                            Example                                                                               Example
                                         2                                                                                   3          2
                            Solve 25tan x −90tanx = −81 on the interval [0,2π].                                   Solve 2sin x +9sin x +4sinx on the interval [0,π].
                            Rewrite as 25tan2x −90tanx +81 = 0 and factor.                                        Common factor, then decompose.
                                               (5tanx −9)2 = 0                                                      (sinx)(2sinx +1)(sinx +4) =0
                                                                    −19
                                                                                 −1        −1 1

                                                            x = tan      5                                                                        x =sin     (0),sin     −2 or
                                                                                                                                                          −1
                            Since tangent is positive in quadrants 1 and 3, x ≈ 1.06 or                                                                sin   (−4)
                            x ≈ π +1.06 ≈ 4.21.                                                                                  −1
                                                                                                                  When x = sin      (0), x = 0 or π, both of which are on [0,π].
                                                                                                                                 −1 1
          7π    11π
                                                                                                                  When x = sin       −2 , x = 6 or 6 , neither of which is on
                                                                                                                                                   −1
                                                                                                                  [0,π]. The case where x = sin       (−4) is impossible.
                                                                                                                  So the only two valid solutions are x = 0 and x = π.
                            J. Garvin — Solving Trigonometric Equations                                           J. Garvin — Solving Trigonometric Equations
                            Slide 7/12                                                                            Slide 8/12
                                                                                       trigonometry                                                                          trigonometry
                            Solving Trigonometric Equations                                                       Solving Trigonometric Equations
                            Example                                                                               Example
                                       2                                                                          Solve 3sin2x −2 = 0 on the interval [0,2π].
                            Solve 3sec x +14secx −5 = 0 on the interval [0,π].
                                 (3secx −1)(secx +5) = 0                                                                                 sin2x = 2
                                                                     
                                                                            3      

                                                                 −1 1           −1                                                                    −1 2
                                                         x = sec      3  or sec    (−5)                                                      2x = sin     3
                                                                                                                  Since sine is positive in quadrants 1 and 2, it appears that
                                                                               −1                                 there are two possible solutions on [0,π].
                            Since secant is the reciprocal of cosine, x = cos     (3)
                                                     −1 1

                            (impossible) or x = cos      −5 .                                                     Since f (x) = 3sin2x − 2 has a period of π, however, a value
                            Cosine is negative in quadrants 2 and 3, but quadrant 3 is                            at x will have an image at x + π, falling on [π,2π].
                            not on [0,π].                                                                         This means that there are four solutions on [0,2π].
                            Thus, the only solution is x ≈ 1.77.
                            J. Garvin — Solving Trigonometric Equations                                           J. Garvin — Solving Trigonometric Equations
                            Slide 9/12                                                                            Slide 10/12
                                                                                       trigonometry                                                                          trigonometry
                            Solving Trigonometric Equations                                                       Questions?
                                            −12

                            When 2x = sin        3  is in quadrant 1, 2x ≈ 0.730, so
                            x ≈ 0.365.          

                                            −1 2
                            When 2x = sin        3  is in quadrant 2,
                            2x ≈ π −0.730 ≈ 2.412, so x ≈ 1.206.
                            If x ≈ 0.365, then its image is at 0.365 + π ≈ 3.507.
                            If x ≈ 1.206, then its image is at 1.206 + π ≈ 4.348.
                            Thus, the four solutions are approximately 0.365, 1.206,
                            3.507 and 4.348.
                            J. Garvin — Solving Trigonometric Equations                                           J. Garvin — Solving Trigonometric Equations
                            Slide 11/12                                                                           Slide 12/12