 SOLVING TRIGONOMETRIC EQUATIONS Note: If you would like a review of trigonometry, click on trigonometry.

Example 2:        Solve for x in the following equation. There are an infinite number of solutions to this problem.

Since denominators of fractions cannot equal zero, real numbers that cause the denominators to equal zero must be eliminated from the set of possible solutions. and Therefore, before we even start to solve the problem, the set of real numbers in the set must be excluded from the possible set of solutions.

To simplify the equation, let's multiple the second fraction by 1 in the form , simplify and solve. The result will be an equation that may not be equivalent to the original equation, but an equation where we can solve for x. With this type of manipulation, there may be extraneous answers. In other words, you may come up with answers for the new equation that are not solutions to the original equation. Therefore, check your answers with the original equation. How do we isolate the x in this equation? We could take the arcsine of both sides of the equation. However, the sine function is not a one-to-one function.

Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The sine function is one-to-one on the interval If we restrict the domain of the sine function to that interval , we can take the arcsine of both sides of each equation. The reference angle is x. We know that Therefore, if then  Since the period of equals , these solutions will repeat every units. The exact solutions are where n is an integer.

The approximate values of these solutions are where n is an integer.

You can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.

You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. The solutions of the original equation are the x-intercepts of this graph.

Algebraic Check:

Check solution Left Side: Right Side: Since the left side of the original equation equals the right side of the original equation when you substitute 0.7297277 for x, then 0.7297277 is a solution.

Check solution Left Side: Right Side: Since the left side of the original equation equals the right side of the original equation when you substitute 2.411865 for x, then 2.411865 is a solution.

We have just verified algebraically that the exact solutions are and and these solutions repeat every units. The approximate values of these solutions are and 2.411865 and these solutions repeat every units.

Graphical Check:

Graph the function , formed by subtracting the right side of the original equation from the left side of the original equation.

Note that the graph crosses the x-axis many times indicating many solutions. Let's check a few of these x-intercepts against the solutions we derived.

Verify the graph crosses the x-axis at 0.7297277. Since the period is , you can verify that the graph also crosses the x-axis again at 0.7297277+6.2831853=7.012913 and at , etc.

Verify the graph crosses the x-axis at 2.411865. Since the period is , you can verify that the graph also crosses the x-axis again at 2.411865+6.2831853=8.69505 and at , etc.

Note: If the problem were to find the solutions in the interval , then you choose those solutions from the set of infinite solutions that belong to the set  and If you would like to work another example, click on Example.

If you would like to test yourself by working some problems similar to this example, click on Problem.

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[Calculus] [Complex Variables] [Matrix Algebra] S.O.S MATHematics home page Author: Nancy Marcus