1. Solving Trigonometric Equations Digital Lesson

2. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 x y 1 -19 π 6 -11 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

3. 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

4. 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

5. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 1 -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

6. 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) =. 9 16 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

7. 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

8. 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 6 1 2 It follows that sin  = - or sin  = -1. 1 2 Quadratic Form

9. 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  = -3 1 3  = sin  1 ( ) = 0.3398 and  = π  sin  1 ( ) = 2.8107. 1 3 1 3 Example: Solutions in an interval

10. 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 (0.6810249 ) = 0.8216349 and  = cos  1 (  0.8810249) = 2.6488206 Therefore, cos  = 0.6810249 or –0.8810249. 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 ± ) = 0.6810249 or -0.8810249 10 Example: Solve quadratic equation

11. 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 45-45-90 triangles. 1 1