SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.7a:         Solve for x in the equation

where n is an integer.

The approximate values of these solutions are

where n is an integer.

Solution:

There are an infinite number of solutions to this problem.

In the equation there are two different trigonometric terms. To make life easier for yourself, convert the equation to an equivalent equation in terms of just sine terms or just cosine terms. Which one shall it be? The easiest way is to convert the trigonometric terms to cosine terms because from the identity , you can solve for . Substitute for in the original equation.

Rewrite the equation in factored form and solve for .

The only way the product of two factors equals zero is if at least one of the factors equals zero. This means that if or

We have transformed a difficult problem into two less difficult problems. To find the solutions to the original equation, , we first find the solutions to in the equations and

and

To solve for x, we must isolate x, and just how do we do that? We could take the arccosine of both sides. However, the cosine function is not a one-to-one function.

We can restrict the domain of the function 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 Therefore, if , then

We complete the problem by solving for the second factor.

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 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 the left side of the original equation and graphing the function formed by the right side of the original equation. The x-coordinates of the points of intersection are the solutions. The right side of the equation is 0 and <tex2htmlcommentmark> f(x)=0 is the x-axis. So really what you are looking for are the x-intercepts to the function formed by the left side of the equation.

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 1.4274488 for x, then 1.4274488 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 -1.4274488 for x, then -1.4274488is 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 0.927295 for x, then 0.927295 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 -0.927295 for x, then -0.927295 is a solution. We have just verified that the exact solutions are and and these solutions repeat every units. The approximate values of these solutions are and and these solutions repeat every units.

Graphical Check:

Graph the equation Note that the graph crosses the x-axis many times indicating many solutions.

Verify the graph crosses the x-axis at -0.927295. Since the period is , the graph also crosses the x-axis again at <tex2htmlcommentmark> -0.927295+6.2831853=5.35589 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> 0.927295+6.2831853=7.21048 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> 1.4274488+6.2831853=7.710634 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> -1.4274488+6.2831853=4.8557365 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

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Author: Nancy Marcus