## SOLVING TRIGONOMETRIC EQUATIONS

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

Example 3:        Solve for x in the following equation.

There are an infinite number of solutions to this problem.

We can make the solution process easier if we convert all the trigonometric terms to like trigonometric terms.

One common trigonometric identity is Since we can solve for let's replace the term in the original equation with The result is an equation equivalent to the original equation with like trigonometric terms, in this case, sine terms.

Isolate the sine term. Since this equation is not easily factored, we can isolate the sine term by solving for using the Quadratic Formula.

We just transformed a difficult problem into two easier problems. To find the solution(s) to the original equation, we find the solution(s) to the equations

To do this we have to isolate the x. How do we isolate the x? 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.

We know that Therefore, if , then

.

We complete the problem by solving the second equation.

The values of range between -1 and +1 and <tex2htmlcommentmark> -1.6407544 is not in this range. Therefore,

Since the period of equals , these solutions will repeat every units. The exact solutions are

and

where n is an integer.

The approximate values of these solutions are

and

where n is an integer.

You can check each solution algebraically by substituting the 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 x=0.900458

Left Side:

Right Side:

Since the left side of the original equation equals the right side of the original equation when you substitute 0.900458 for x, then 0.900458 is a solution.

Check solution x=2.241134

Left Side:

Right Side:

Since the left side of the original equation equals the right side of the original equation when you substitute 2.241134 for x, then 2.241134 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 2.241134 and these solutions repeat every units.

Graphical Check:

Graph the equation Note 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 0.900458. Since the period is , you can verify that the graph also crosses the x-axis again at 0.900458+6.2831853=7.1836433 and at , etc.

Verify the graph crosses the x-axis at Since the period is , the graph also crosses the x-axis again at <tex2htmlcommentmark> 2.241134+6.2831853=8.52432 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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[Algebra] [Trigonometry]
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Author: Nancy Marcus