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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
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Solving Trigonometric Equations
Unit 6 continued Book sections
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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 = k π and x = k π (k = 0, ±1, ± 2, ± 3, ). 6 π
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Example: General Solution
Find the general solution for the equation sec = 2. From cos = , it follows that cos = . 1 sec cos( kπ) = 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 = ± kπ . π 3 Example: General Solution
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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.
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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
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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
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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: = kπ and = kπ, from sin = - π 6 7π 1 2 = -π + 2kπ, from sin = -1
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Worksheet 1 – 3, 9 – 17 odd, 64 – 70 even Assignment
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