University:
Carleton College
Course:
MATH 111 | Introduction to Calculus
Academic year:

2024
Views:

135

Pages:

4
Author:

Clint W.

HPC Section 5.5 D2 Notes Name: Warm Up: Solve the equation in the interval [0, 2π) a. √2 sin x + 1 = 0 sin x = -1/√2 x = 5π/4, 7π/4 b. 3cot²x - 1 = 0 cot²x = 1/3 cot x = ±1/√3 x = π/3, 2π/3, 4π/3, 5π/3 c. 2sin²x = 2 + cos x 2(1-cos²x) = 2 + cos x 2 - 2cos²x = 2 + cos x 0 = 2cos²x + cos x 0 = cos x(2cos x + 1) cos x = 0 or cos x = -1/2 x = π/2, 3π/2, 2π/3, 4π/3 Finding the "general solution" of a trigonometric equation: * Finds ALL solutions, not just the ones between 0 and 2π * Find these by adding a multiple of the period of the function (coterminal angles) * Sometimes answers can be "condensed" or simplified (when radians apart are π radians apart) Example 1: Find the "general solution" for parts a-c of the warm up a. 5π/4 + 2nπ 7π/4 + 2nπ b. π/3 + nπ 2π/3 + nπ c. π/2 + nπ 2π/3 + 2nπ 4π/3 + 2nπ Example 2: Solve the equation. (When you see these directions, it means "find the general solution," not just the solutions in the interval [0, 2π).) a. sin 2x = -cos x 2sin x cos x + cos x = 0 cos x(2sin x + 1) = 0 cos x = 0 or sin x = -1/2 x = π/2 + nπ, 7π/6 + 2nπ, 11π/6 + 2nπ b. cos 2x + sin x = 0 1 - 2sin²x + sin x = 0 0 = 2sin²x - sin x - 1 0 = (2sin x + 1)(sin x - 1) sin x = -1/2 or sin x = 1 x = 7π/6 + 2nπ, 11π/6 + 2nπ, π/2 + 2nπ c. tan²x - 1 = 0 tan²x = 1 tan x = ±1 x = π/4 + nπ/2 Using the "general solution" to solve multiple angles Example 3: Solve the equation. (We will NOT be using double-angle identities to solve parts a and b! We will use "u substitutions" and the "general solution") a. sin 2x = 0 Let u = 2x sin u = 0 u = 0π + nπ 2x = 0π + nπ x = 0π + nπ/2 b. cos 2x = 0 Let u = 2x cos u = 0 u = π/2 + nπ 2x = π/2 + nπ x = π/4 + nπ/2 c. 2cos 3x - 1 = 0 Let u = 3x 2cos u - 1 = 0 cos u = 1/2 u = π/3 + 2nπ u = 5π/3 + 2nπ 3x = π/3 + 2nπ 3x = 5π/3 + 2nπ x = π/9 + 2nπ/3 x = 5π/9 + 2nπ/3 d. sin 2x - √3/2 = 0 Let u = 2x sin u = √3/2 u = π/3 + 2nπ u = 2π/3 + 2nπ 2x = π/3 + 2nπ 2x = 2π/3 + 2nπ x = π/6 + nπ x = π/3 + nπ Example 4: Solve the equation in the interval [0, 2π) a. 2cos 3x - 1 = 0 x = π/9 + 2nπ/3 x = 5π/9 + 2nπ/3 n = 0: π/9, 5π/9 n = 1: 7π/9, 13π/9 n = 2: Can't go over n=1 or answers will be greater than 2π b. sin 2x - √3/2 = 0 x = π/6 + nπ x = π/3 + nπ n = 0: π/6, π/3 n = 1: 7π/6, 4π/3 n = 2: Can't go over n=1 or answers will be greater than 2π HPC 5.5 D2 Homework Worksheet Solve the equation. 1. tan x + √3 = 0 tan x = -√3 x = 2π/3 + nπ 2. sec x csc x = 2 csc x sec x csc x - 2 csc x = 0 csc x(sec x - 2) = 0 csc x = 0 or sec x = 2 sin x = ∞ or cos x = 1/2 x = π/3 + 2nπ, 5π/3 + 2nπ 3. 2cos²x = 2 + sin x 2(1-sin²x) = 2 + sin x 2 - 2sin²x = 2 + sin x 0 = 2sin²x + sin x 0 = sin x(2sin x + 1) sin x = 0 or sin x = -1/2 x = 0 + nπ, 7π/6 + 2nπ, 11π/6 + 2nπ 4. csc²x - 2 = 0 csc²x = 2 sin²x = 1/2 sin x = ±√(1/2) = ±1/√2 x = π/4 + nπ/2 5. 3tan³x = tan x 3tan³x - tan x = 0 tan x(3tan²x - 1) = 0 tan x = 0 or tan²x = 1/3 tan x = ±1/√3 x = 0 + nπ, π/6 + nπ, 5π/6 + nπ 6. sin²x + cos x + 1 = 0 1 - cos²x + cos x + 1 = 0 0 = cos²x - cos x - 2 0 = (cos x - 2)(cos x + 1) cos x = 2 (no solution) cos x = -1 x = π + nπ