SOLVING TRIGONOMETRIC EQUATIONS


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


Example 3:        Solve for x in the following equation.



\begin{displaymath}\displaystyle \frac{1+\sin x}{\cos x}+\displaystyle \frac{\cos x}{1+\sin x}=4\sec x\end{displaymath}

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. $\cos x\neq 0\rightarrow x\neq \pm \displaystyle \frac{\pi }{2}\pm n\pi
,\qquad $and         $\sin x\neq -1\rightarrow x\neq \displaystyle \frac{3\pi }{2}\pm 2n\pi
.$ Therefore, before we even start to solve the problem, the set of real numbers in the set $\left\{ \pm \displaystyle \frac{\pi }{2}\pm n\pi \right\} $ must be excluded from the possible set of solutions.



The equation has three different trigonometric terms. Let's manipulate the equation so that we have an equation of like trigonometric terms.



First, let's multiple the second fraction by 1 in the form $\displaystyle \frac{1-\sin x}{%
1-\sin x}$, simplify and solve. The result will be an equation that is not equivalent to the original equation, but an equation that 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.

\begin{displaymath}\begin{array}{rclll}
&& \\
\displaystyle \frac{1+\sin x}{\co...
...cos x} \\
&& \\
2 &\neq &4\rightarrow \phi \\
&&
\end{array}\end{displaymath}

There are no solution.




If you made a mistake as you went through this kind of exclusion analysis, you can catch your mistake in the check. 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. If there are not x-intercepts, there are no real solutions.


Algebraic Check:


Choose any real number, say 252, to illustrate that no solution checks.


Left Side:

\begin{displaymath}\begin{array}{rclll}
\displaystyle \frac{1+\sin x}{\cos x}+\d...
... }{1+\sin \left( 252\right) }\approx 2.55685 \\
&&
\end{array}\end{displaymath}

Right Side:         $4\sec x=\displaystyle \frac{4}{\cos x}\approx \displaystyle \frac{4}{\cos \left(
252\right) }\approx 5.11370247d\bigskip $

Since the left side of the equation does not equal the right side of the equation after the substitution of 252 for x, then x=252 is not a solution.




We have just verified algebraically that there are no solutions to the original problem.




Graphical Check:


Graph the equation $f(x)=\displaystyle \frac{1+\sin x}{\cos x}+\displaystyle \frac{\cos x}{1+\sin x}%
-4\sec x.$ Note that the graph never crosses the x-axis indicating no real solutions.



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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Author: Nancy Marcus

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