Note: If you would like a review of trigonometry, click on trigonometry.
Example 4: Solve for x in the following equation.
There are an infinite number of solutions to this problem.
We can make the solution easier if we convert all the trigonometric terms to like trigonometric terms.
We know that
in the original equation with
and multiply both sides of the equation by
we will have an equation with cosine trigonometric.
Isolate the cosine term. Since this equation is not easily factored, we can isolate the cosine term by solving for using the Quadratic Formula.
We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, we find the solutions to the equations
To do this we have to isolate the x. How do we isolate the x? We could take the arccosine of both sides. However, the cosine 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 cosine function is one-to-one on the interval If we restrict the domain of the cosine function to that interval , we can take the arccosine of both sides of each equation.
We know that
We complete the problem by solving the second equation.
The values of range between -1 and +1 and <tex2htmlcommentmark> -1.7836116 is not in this range. Therefore,
Since the period of
these solutions will repeat
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 ethe 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.
Check solution x=0.8753157
Since the left side of the original equation equals the right side of the original equation when you substitute 0.8753157 for x, then 0.8753157 is a solution.
Check solution x=-0.8753157
Since the left side of the original equation equals the right side of the original equation when you substitute -0.8753157 for x, then -0.8753157is a solution.
We have just verified algebraically that the exact solutions are
Graph the equation If your graphing program does not have a secant function, graph 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 0.8753157. Since the period is , you can verify that the graph also crosses the x-axis again at 0.8753157+6.2831853=7.158501 and at , etc.
Verify the graph crosses the x-axis at -0.8753157 Since the period is , the graph also crosses the x-axis again at <tex2htmlcommentmark> -0.8753157+6.2831853=5.4078696 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 test yourself by working some problems similar to this example, click on Problem.
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