MTH 112 Elementary Functions Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Section 5 Solving Trigonometric Equations

Trigonometric Equations An equation that involves a trigonometric function of a variable or an expression involving a variable. –Identity True for all values in the domain of the variable. –Conditional True for some values in the domain of the variable. –Fallacy True for none of the values in the domain of the variable.

Simple Trigonometric Equations cos x = a where: -1  a  1 x = cos -1 a gives a solution between 0 and . x = −cos -1 a is a 2nd solution (symmetry of the unit circle) Add 2k  for the coterminal solutions (k is an integer) a − cos -1 a + 2k  cos -1 a Use values of k that give solutions within the specified range.

Simple Trigonometric Equations sin x = b where: -1  b  1 x = sin -1 b gives a solution between −  /2 and  /2. x =  −sin -1 b is a 2nd solution (symmetry of the unit circle) Add 2k  for the coterminal solutions (k is an integer) b + 2k  Use values of k that give solutions within the specified range. sin -1 b  − sin -1 b

Simple Trigonometric Equations tan x = c where: -   c   x = tan -1 c gives a solution between −  /2 and  /2. x =  +tan -1 c is a 2nd solution (symmetry of the unit circle) Add 2k  for the coterminal solutions (k is an integer) Use values of k that give solutions within the specified range. c tan -1 c  + tan -1 c + 2k 

Solving Trigonometric Equations Strategy … 1.Use only sin x, cos x, and/or tan x. –If possible, use only one of these function. 2.Algebraically solve for the trigonometric function(s). –You must isolate the/each function: trig(x) = n 3.Solve the resulting simple trigonometric equation(s).

Solving Trigonometric Equations with a function as an argument. Strategy … 1.Replace the argument with a variable (say: u). ALL 2.Solve the equation (as before) for u. Find ALL solutions. 3.In the solutions, replace u with the argument. 4.Solve for the variable and simplify (if possible). Example: Argument is a function. 2 sin u = 1 sin u = ½  u =  /6 + 2k  and u = 5  /6 + 2k  3x − 1 =  /6 + 2k  and 3x − 1 = 5  /6 + 2k  x = (  /6 + 2k  + 1)/3 and x = (5  /6 + 2k  + 1)/3 Use values of k that give solutions within the specified range.