Solving Trigonometric Equations Digital Lesson

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 x y π π 6 -7 π 6 π 6 5 π 6 13 π 6 17 π 6 25 π 6 -π-π -2 π -3 π π 2π2π 3π3π 4π4π All the solutions for x can be expressed in the form of a general solution. x = is one of infinitely many solutions of y = sin x. π 6 x = + 2k π and x = 5 + 2k π (k = 0, ±1, ± 2, ± 3,  ). 6 ππ 6 sin x = is a trigonometric equation. y = y=sin x

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 Find the general solution for the equation sec  = 2. All values of  for which cos  = are solutions of the equation. Two solutions are  = ±. All angles that are coterminal with ± are also solutions and can be expressed by adding integer multiples of 2π. π 3 π 3 The general solution can be written as  = ± + 2kπ. π 3 From cos  =, it follows that cos  =. 1 sec  Example: General Solution cos( + 2kπ) = π 3 -π-π 3 x y Q 1 P

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 Example: Solve tan x = 1. The graph of y = 1 intersects the graph of y = tan x infinitely many times. General solution: x = + kπ for k any integer. π 4 Points of intersection are at x = and every multiple of π added or subtracted from. π 4 π 4 Example: Solve tan x=1 y 2 x -π-ππ2π2π3π3π x = -3π 2 y = tan(x) x = -π 2 x = π x = 3π 2 x = 5π 2 y = 1 - π – 2π 4 - π – π 4 π 4 π + π 4 π + 2π 4 π + 3π 4

Copyright © by Houghton Mifflin Company, Inc. All rights reserved π-π 4 Example: Solve the equation 3sin x + = sin x for  ≤ x ≤. π 2 π 2 2sin x + = 0 Collect like terms. 3sin x  sin x + = 0 3sin x + = sin x sin x =  1 x y y = - x =  is the only solution in the interval  ≤ x ≤. π 2 π 2 π 4 Example: Solve the Equation

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 x y Take the fourth root of both sides to obtain: cos(2x)= ± From the unit circle, the solutions for 2  are 2  = ± + kπ, k any integer. π 6 Example: To find all solutions of cos 4 (2x) = Answer:  = ± + k ( ), for k any integer. 12 π 2 π Example: Find all solutions using unit circle 1 π 6 -π-π 6 x = - x = π π

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 Find all solutions of the trigonometric equation: tan 2  + tan  = 0. The solutions for tan  = 0 are the values  = kπ, for k any integer. Therefore, tan  = 0 or tan  = -1. tan 2  + tan  = 0 Original equation tan  (tan  +1) = 0Factor. The solutions for tan  = 1 are  = - + kπ, for k any integer. π 4 Example: Find all solutions

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 The trigonometric equation 2 sin 2  + 3 sin  + 1 = 0 is quadratic in form. 2 sin 2  + 3 sin  + 1 = 0 implies that x  = -π + 2kπ, from sin  = -1 (2 sin  + 1)(sin  + 1) = 0. Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0. Solutions:  = - + 2kπ and  = + 2kπ, from sin  = - π 6 7π7π It follows that sin  = - or sin  = Quadratic Form

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 8 sin  = 3(1  sin 2  ) Use the Pythagorean Identity. Rewrite the equation in terms of only one trigonometric function. Example: Solve 8 sin  = 3 cos 2  with  in the interval [0, 2π]. 3 sin 2  + 8 sin   3 = 0. A “quadratic” equation with sin x as the variable Therefore, 3 sin   1 = 0 or sin  + 3 = 0 (3 sin   1)(sin  + 3) = 0 Factor. s Solutions: sin  = or sin  =  = sin  1 ( ) = and  = π  sin  1 ( ) = Example: Solutions in an interval

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 Solve: 5cos 2  + cos  – 3 = 0 for 0 ≤  ≤ π. This is the range of the inverse cosine function. The solutions are:  = cos  1 ( ) = and  = cos  1 (  ) = Therefore, cos  = or – Use the calculator to find values of  in 0 ≤  ≤ π. The equation is quadratic. Let y = cos  and solve 5y 2 + y  3 = 0. y = (-1 ± ) = or Example: Solve quadratic equation

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Example: Find the intersection points of the graphs of y = sin  and y = cos . x y The general solution is  = + kπ, for k any integer. π 4 Example: Find points of intersection The two solutions for  between 0 and 2π are and. 5π5π 4 π 4 5 -π-π 4 -π-π 4 π 4 + kπ π 4 The graphs of y = sin  and y = cos  intersect at points where sin  = cos . This is true only for triangles. 1 1

Try to get equations in terms of one trig function by using identities. Try to get trig functions of the same angle. If one term is cos2  and another is cos  for example, use the double angle formula to express first term in terms of just  instead of 2  Get one side equals zero and factor out any common trig functions See if equation is quadratic in form and will factor. (replace the trig function with x to see how it factors if that helps) If the angle you are solving for is a multiple of , don't forget to add 2  to your answer for each multiple of  since  will still be less than 2  when solved for. HELPFUL HINTS FOR SOLVING TRIGONOMETRIC EQUATIONS Be on the look-out for ways to substitute using identities

Use a graphing utility to solve the equation. Express any solutions rounded to two decimal places. Graph this side as y 1 in your calculator Graph this side as y 2 in your calculator You want to know where they are equal. That would be where their graphs intersect. You can use the trace feature or the intersect feature to find this (or these) points (there could be more than one point of intersection). There are some equations that can't be solved by hand and we must use a some kind of technology.

This was graphed on the computer with graphcalc, a free graphing utility you can download at After seeing the initial graph, lets change the window to get a better view of the intersection point and then we'll do a trace. Rounded to 2 decimal places, the point of intersection is x = 0.53 check: This is off a little due to the fact we approximated. If you carried it to more decimal places you'd have more accuracy.

Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating some of this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar

Solve the equation: 2 cos 2 x  cos x  1  0, 0  x  2 . The solutions in the interval [0, 2  ) are  / 3, , and 5  / 3. Solution The given equation is in quadratic form 2t 2  t  1  0 with t  cos x. Let us attempt to solve the equation using factoring. 2 cos 2 x  cos x  1  0 This is the given equation. (2 cos x  1)(cos x  1)  0 Factor. Notice that 2t 2 + t – 1 factors as (2t – 1)(2t + 1). cos x  1/2 2 cos x  1 cos x  1 Solve for cos x. 2 cos x  1  0 or cos x  1  0 Set each factor equal to 0. Text Example x   x  2  x  

Example Solve the following equation: Solution:

Example Solve the equation on the interval [0,2  ) Solution:

Example Solve the equation on the interval [0,2  ) Solution:

Example Solve the equation on the interval [0,2  ) Solution: