Warm Up Take out a copy of the unit circle Find where sin θ = - ½ from 0 to 2π Find where sin θ = - ½ for all values of θ Factor x2 + x2y2 Warm Up

Solving Trigonometric Equations Unit 6 continued Book sections 5.3-5.5

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

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 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 the equation 3sin x + = sin x for  ≤ x ≤ . π 2 3sin x + = sin x 1 x y y = - 3sin x  sin x + = 0 2sin x + = 0 Collect like terms. 1 -π 4 sin x =  x =  is the only solution in the interval  ≤ x ≤ . π 2 4

Find all solutions of the trigonometric equation: 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

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

Worksheet 1 – 3, 9 – 17 odd, 64 – 70 even Assignment