1. 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, 2  or 360° for sin and cos, and  or 180° for tan.

2. 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°.

3. 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) = 0Factor 2 cos x - 1 = 0 2 cos x = 1 cos x = ½ x = 60° orcos x – 2 = 0 cos x = 2 There is no solution for cos x = 2 since –1 < cos x < 1

4. 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 2 tan x sin x + 2 sin x – tan x – 1 = 0 Subtract tan x + 1 from both sides. (tan x + 1) (2 sin x – 1) = 0 factor tan x + 1 = 0 tan x = -1 or2 sin x – 1 = 0 2 sin x = 1 sin x = ½ When all the values of x are required, the solution should be represented as x + 360k° for sin and cos, and x + 180k° for tan, where k is any integer. x = -45° + 180k° x = 30° + 360k° or x = 150° + 360k°

5. 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 cos² x – cos x = 0 1 – sin² x = cos² x cos x (cos x – 1) = 0 Factor cos x = 0 x = 90° or cos x – 1 = 0 cos x = 1 x = 0° So the solutions are 0° and 90°

6. 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 2 sin x = 0 sin x = 0 x = 0° or 180º or 1 + sin x = 0 sin x = -1 x = 270º

7. It’s important to always check your solutions. Some may not actually be solutions to the original equation. x = 0º or 180ºx = 270º cos x = 1 + sin x cos 0º = 1 + sin 0° 1 = 1 cos 180º = 1 + sin 180º -1 = 1 + 0 -1 ≠ 1 ☻ cos 270º = 1 + sin 270º 0 = 1 + (-1) 0 = 0☻ Based on the check, 180º is not a solution

8. Assignment  Page 390 –# 17 - 38