SOLVING TRIGONOMETRIC EQUATIONS

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

Example 2:        Solve for x in the following equation.

$$5\cos ^{2}\left( x\right) -6\sin \left( x\right) =0$$

There are an infinite number of solutions to this problem.

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

One common trigonometric identity is $\sin ^{2}(x)+\cos ^{2}(x)=1$ and $\cos ^{2}\left( x\right) =1-\sin ^{2}\left( x\right) .$

Replace the $\cos ^{2}\left( x\right)$ in the original equation with $% 1-\sin ^{2}\left( x\right)$ and we have an equivalent equation with sines terms.

$$\begin{array}{rclll} && \\ 5\cos ^{2}\left( x\right) -6\sin ... ...eft( x\right) +6\sin \left( x\right) -5 &=&0 \\ && \end{array}$$

Isolate the sine term. Since this equation is not easily factored, we can solve it by using the quadratic formula to solve for $\sin \left( x\right) .$

$$\begin{array}{rclll} && \\ 5\sin ^{2}\left( x\right) +6\sin ... ...isplaystyle \frac{-3\pm \sqrt{34}}{5} \\ && \\ && \end{array}$$

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 $\sin \left( x\right) =\displaystyle \displaystyle \frac{-3\pm \sqrt{34}}{5}.\bigskip \bigskip$

How do we isolate the x? We could take the arcsine of both sides. 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 graph of the sine function is one-to-one on the interval $\left[ -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{2}% \right] .$ If we restrict the domain of the sine function to that interval , we can take the arcsine of both sides of each equation.

$$\begin{array}{rclll} (1)\qquad \sin \left( x\right) &=&\displ... ...sqrt{34}}{5}\right) \approx 0.6018767 \\ && \\ && \end{array}$$

We know that $\sin \left( x\right) =\sin \left( \pi -x\right) .$ Therefore, if $\sin (x)=\displaystyle \displaystyle \frac{-3+\sqrt{34}}{5}$, then $\sin (\pi -x)=\displaystyle \displaystyle \frac{-3+\sqrt{34% }}{5}.$

$$\begin{array}{rclll} && \\ \left( 2\right) \qquad \sin (\pi ... ...qrt{34}}{5}\right) \approx 2.53971596 \\ && \\ && \end{array}$$

We complete the problem by solving the second equation.

$$\begin{array}{rclll} && \\ (3)\qquad \sin \left( x\right) &=... ...rt{34}}{5}\approx -1.7661904\rightarrow \phi \\ && \end{array}$$

The values of $\sin \left( x\right)$ range between -1 and +1 and <tex2html_comment_mark> -1.7661904 is not in this range. Therefore, $\sin \left( x\right) \neq \displaystyle \displaystyle \frac{-3-\sqrt{34}}{5}.\bigskip\bigskip\bigskip$

Since the period of $\sin (x)$ equals $2\pi$, these solutions will repeat every $2\pi$ units. The exact solutions are

$$\begin{array}{rclll} x_{1} &=&\sin ^{-1}\left( \displaystyle ... ...frac{-3+\sqrt{34}}{5}\right) \pm 2\pi \\ && \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.6018767\pm 6.28... ...2} &\approx &2.53971596\pm 6.2831853n \\ && \\ && \end{array}$$

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 x=0.6018767

Left Side:

$$5\cos ^{2}\left( x\right) -6\sin \left( x\right) \approx 5\co... ...6018767\right) -6\sin \left( 0.6018767\right) \approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.6018767 for x, then 0.6018767 is a solution.

Check solution x=2.53971596

Left Side:

$$5\cos ^{2}\left( x\right) -6\sin \left( x\right) \approx 5\co... ...971596\right) -6\sin \left( 2.53971596\right) \approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 2.53971596 for x, then 2.53971596is a solution.

We have just verified algebraically that the exact solutions are $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{-3+\sqrt{34}}{5}\right) \pm 2\pi$ and $x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{-3+\sqrt{34}}{5}\right) \pm 2\pi$ and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $x\approx 0.6018767$ and 2.53971596 and these solutions repeat every $% \pm 6.2831853$ units.

Graphical Check:

Graph the equation $f(x)=5\cos ^{2}\left( x\right) -6\sin \left( x\right) .$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.

The graph crosses the x-axis at 0.6018767. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 0.6018767+6.2831853=6.885062 and at $0.6018767+2\left( 6.2831853\right) =13.168247$, etc.

The graph crosses the x-axis at $\ 2.53971596$ Since the period is $2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2html_comment_mark> 2.53971596+6.2831853=8.822901 and at $2.53971596+2\left( 6.2831853\right) =15.10608657$, etc.$.\bigskip$

Note: If the problem were to find the solutions in the interval $\left[ 0,2\pi \right]$, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right] :$ $x\approx 0.6018767$ and $2.53971596.\bigskip\bigskip\bigskip\bigskip$

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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