Identities and Conditional equations

Trigonometric equations can be broken into two categories: identities and conditional equations. Identities are true for any angle, whereas conditional equations are true only for certain angles. Identities can be tested, checked, and created using knowledge of the eight fundamental identities. We already discussed these processes in Trigonometric Identities . The following sections are dedicated to explaining how to solve conditional equations.

Conditional Equations

When solving a conditional equation, a general rule applies: if there is one solution, then there are an infinite number of solutions. This strange truth results from the fact that the trigonometric functions are periodic, repeating every 360 degrees or 2Π radians. For example, the values of the trigonometric functions at 10 degrees are the same as they are at 370 degrees and 730 degrees. The form for any answer to a conditional equation is θ +2, where θ is one solution to the equation, and n is an integer. The shorter and more common way to express the solution to a conditional equation is to include all the solutions to the equation that fall within the bounds [0, 2Π), and to omit the "+2" part of the solution. since it is assumed as part of the solution to any trigonometric equation. Because the set of values from 0 to 2Π contains the domain for all six trigonometric functions, if there is no solution to an equation between these bounds, then no solution exists.

Solutions for trigonometric equations follow no standard procedure, but there are a number of techniques that may help in finding a solution. These techniques are essentially the same as those used in solving algebraic equations, only now we are manipulating trigonometric functions: we can factor an expression to get different, more understandable expressions, we can multiply or divide through by a scalar, we can square or take the square root of both sides of an equation, etc. Also, using the eight fundamental identities, we can substitute certain functions for others, or break a functions down into two different ones, like expressing tangent using sine and cosine. In the problems below, we'll see just how helpful some of these techniques can be.


2 cos(x) - 1 = 0    

2 cos(x) = 1    

cos(x) =    

x = ,    

In this problem, we came up with two solutions in the range [0, 2Π) : x = , and x = . By adding 2 to either of these solutions, where n is an integer, we could have an infinite number of solutions.


sin(x) = 2 cos2(x) - 1    

sin(x) = 2(1 - sin2(x)) - 1    

sin(x) = 1 - 2 sin2(x)    

2 sin2(x) + sin(x) - 1 = 0    

(sin(x) + 1)(2 sin(x) - 1) = 0    

At this point, after factoring, we have two equations we need to deal with separately. First, we'll solve (sin(x) + 1) = 0, and then we'll solve (2 sin(x) - 1) = 0


sin(x) + 1 = 0    

sin(x) = - 1    

x =    

2 sin(x) - 1 = 0    

sin(x) =    

x = ,    

For the problem, then, we have three solutions: x = ,,. All of them check. Here is one more problem.


sec2(x) + cos2(x) = 2    

1 + tan2(x) + 1 - sin2(x) = 2    

tan2(x) = sin2(x)    

= sin2(x)    

cos2(x) = 1    

cos(x) = ±1    

x = 0, Π