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Solving Trigonometric Equations Trig identities are true for all values of the variable for which the variable is defined. However, trig equations, like algebraic equations, are true for some but not all values of the variable. Trig equations do not have unique solutions. Trig equations have infinitely many solutions. They differ by the period of the function, 2p or 360° for sin and cos, and p or 180° for tan. Solving Trig Equations Trig equations will have unique solutions if the value of the function is restricted to two adjacent quadrants. These solutions are called the principal values. For sin x and tan x, the principal values are in Quadrants I or IV. x is in the interval -90° < x < 90°. For cos x, the principal values are in Q I or II. So x is in the interval 0° < x < 180°. Solve 2 cos² x – 5 cos x + 2 = 0 for the principal values of x. 2 cos² x – 5 cos x + 2 = 0 (2 cos x - 1) (cos x – 2) = 0 2 cos x - 1 = 0 or Factor cos x – 2 = 0 2 cos x = 1 cos x = 2 cos x = ½ There is no solution for cos x = 2 since –1 < cos x < 1 x = 60° Solve 2 tan x sin x + 2 sin x = tan x + 1 for all values of x. 2 tan x sin x + 2 sin x = tan x + 1 Subtract tan x + 1 from 2 tan x sin x + 2 sin x – tan x – 1 = 0 both sides. (tan x + 1) (2 sin x – 1) = 0 factor When all the values tan x + 1 = 0 or 2 sin x – 1 = 0 of x are required, the solution should 2 sin x = 1 tan x = -1 be represented as sin x = ½ x + 360k° for sin x = -45° + 180k° x = 30° + 360k° and cos, and x + 180k° for tan, or x = 150° + 360k° where k is any integer. Solve sin² x + cos 2x – cos x = 0 for the principal values of x. sin² x + cos 2x – cos x = 0 sin² x + (1 – 2 sin² x) – cos x = 0 Express cos 2x in terms of sin x. 1 – sin² x – cos x = 0 Combine like terms 1 – sin² x = cos² x cos² x – cos x = 0 Factor cos x (cos x – 1) = 0 So the solutions or cos x = 0 cos x – 1 = 0 are 0° and 90° cos x = 1 x = 90° x = 0° Solve cos x = 1 + sin x for 0° < x < 360° cos x = 1 + sin x cos² x = (1 + sin x)² Square both sides cos² x = 1 + 2 sin x + sin² x Expand the binomial squared 1 – sin² x = 1 + 2 sin x + sin² x cos² x = 1 – sin² x 0 = 2 sin x + 2 sin² x 0 = 2 sin x (1 + sin x) Factor or 2 sin x = 0 1 + sin x = 0 sin x = -1 sin x = 0 x = 270º x = 0° or 180º It’s important to always check your solutions. Some may not actually be solutions to the original equation. x = 0º or 180º x = 270º cos x = 1 + sin x cos 0º = 1 + sin 0° cos 180º = 1 + sin 180º 1=1 ☻ -1 = 1 + 0 -1 ≠ 1 cos 270º = 1 + sin 270º Based on the check, 0 = 1 + (-1) 180º is not a solution 0=0 ☻ Assignment Page 390 – # 17 - 38