Solving Trigonometric Equations Digital Lesson Solving Trigonometric Equations

sin x = is a trigonometric equation. x = is one of infinitely many solutions of y = sin x. π 6 -1 x y 1 -19π 6 -11π -7π π 5π 13π 17π 25π y = -π -2π -3π π 2π 3π 4π All the solutions for x can be expressed in the form of a general solution. x = + 2k π and x = 5 + 2k π (k = 0, ±1, ± 2, ± 3,  ). 6 π Copyright © by Houghton Mifflin Company, Inc. All rights reserved. y=sin x

Example: General Solution Find the general solution for the equation sec  = 2. From cos  = , it follows that cos  = . 1 sec  cos( + 2kπ) = π 3 -π x y Q 1 P 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 The general solution can be written as  = ± + 2kπ . π 3 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: General Solution

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

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

Example: Find all solutions using unit circle Example: To find all solutions of cos4(2x) = . 9 16 Take the fourth root of both sides to obtain: cos(2x)= ± x y From the unit circle, the solutions for 2 are 2 = ± + kπ, k any integer. π 6 π 1 π 6 -π π x = - x = Answer:  = ± + k ( ), for k any integer. 12 π 2 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Find all solutions using unit circle

Example: Find all solutions Find all solutions of the trigonometric equation: tan2  + tan  = 0. tan2  + tan  = 0 Original equation tan  (tan  +1) = 0 Factor. Therefore, tan  = 0 or tan  = -1. The solutions for tan  = 0 are the values  = kπ, for k any integer. The solutions for tan  = 1 are  = - + kπ, for k any integer. π 4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Find all solutions

2 sin2  + 3 sin  + 1 = 0 implies that The trigonometric equation 2 sin2  + 3 sin  + 1 = 0 is quadratic in form. 2 sin2  + 3 sin  + 1 = 0 implies that (2 sin  + 1)(sin  + 1) = 0. Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0. It follows that sin  = - or sin  = -1. 1 2 Solutions:  = - + 2kπ and  = + 2kπ, from sin  = - π 6 7π 1 2  = -π + 2kπ, from sin  = -1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. x Quadratic Form

Example: Solutions in an interval Example: Solve 8 sin  = 3 cos2  with  in the interval [0, 2π]. Rewrite the equation in terms of only one trigonometric function. 8 sin  = 3(1 sin2  ) Use the Pythagorean Identity. 3 sin2  + 8 sin   3 = 0. A “quadratic” equation with sin x as the variable (3 sin   1)(sin  + 3) = 0 Factor. Therefore, 3 sin   1 = 0 or sin  + 3 = 0 Solutions: sin  = or sin  = -3 1 3  = sin1( ) = 0.3398 and  = π  sin1( ) = 2.8107. 1 3 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. s Example: Solutions in an interval

Example: Solve quadratic equation Solve: 5cos2  + cos  – 3 = 0 for 0 ≤  ≤ π. The equation is quadratic. Let y = cos  and solve 5y2 + y  3 = 0. y = (-1 ± ) = 0.6810249 or -0.8810249 10 Therefore, cos  = 0.6810249 or –0.8810249. Use the calculator to find values of  in 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 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve quadratic equation

Example: Find points of intersection Example: Find the intersection points of the graphs of y = sin  and y = cos . x y π 4 + kπ 1 The two solutions for  between 0 and 2π are and . 5π 4 π 5 -π 4 π 4 + kπ The graphs of y = sin  and y = cos  intersect at points where sin  = cos . This is true only for 45-45-90 triangles. The general solution is  = + kπ, for k any integer. π 4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Find points of intersection