# 7.5 Solving trigonometric equations  (Page 3/7)

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Find all solutions for $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=\sqrt{3}.$

$\frac{\pi }{3}±\pi k$

## Identify all solutions to the equation involving tangent

Identify all exact solutions to the equation $\text{\hspace{0.17em}}2\left(\mathrm{tan}\text{\hspace{0.17em}}x+3\right)=5+\mathrm{tan}\text{\hspace{0.17em}}x,0\le x<2\pi .$

We can solve this equation using only algebra. Isolate the expression $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ on the left side of the equals sign.

$\begin{array}{cc}\hfill 2\left(\mathrm{tan}x\right)+2\left(3\right)& =5+\mathrm{tan}x\hfill \\ \hfill 2\mathrm{tan}\text{\hspace{0.17em}}x+6& =5+\mathrm{tan}\text{\hspace{0.17em}}x\hfill \\ \hfill \text{}2\mathrm{tan}x-\mathrm{tan}x& =5-6\hfill \\ \hfill \mathrm{tan}x& =-1\hfill \end{array}$

There are two angles on the unit circle that have a tangent value of $\text{\hspace{0.17em}}-1:\theta =\frac{3\pi }{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta =\frac{7\pi }{4}.$

## Solve trigonometric equations using a calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

## Using a calculator to solve a trigonometric equation involving sine

Use a calculator to solve the equation $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =0.8,$ where $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in radians.

Make sure mode is set to radians. To find $\text{\hspace{0.17em}}\theta ,$ use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\text{\hspace{0.17em}}$ function. What is shown on the screen is ${\mathrm{sin}}^{-1}\left(\text{\hspace{0.17em}}.$ The calculator is ready for the input within the parentheses. For this problem, we enter $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(0.8\right),\text{\hspace{0.17em}}$ and press ENTER. Thus, to four decimals places,

${\mathrm{sin}}^{-1}\left(0.8\right)\approx 0.9273$

The solution is

$0.9273±2\pi k$

The angle measurement in degrees is

## Using a calculator to solve a trigonometric equation involving secant

Use a calculator to solve the equation $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}\theta =-4,$ giving your answer in radians.

We can begin with some algebra.

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}\theta =-4\\ \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }=-4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{1}{4}\end{array}$

Check that the MODE is in radians. Now use the inverse cosine function.

Since $\text{\hspace{0.17em}}\frac{\pi }{2}\approx 1.57\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\pi \approx 3.14,$ 1.8235 is between these two numbers, thus $\text{\hspace{0.17em}}\theta \approx \text{1}\text{.8235}\text{\hspace{0.17em}}$ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See [link] .

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is $\text{\hspace{0.17em}}\theta \text{​}\text{​}\text{'}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}\text{\hspace{0.17em}}$ The other solution in quadrant III is $\text{\hspace{0.17em}}\pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}$

The solutions are $\text{\hspace{0.17em}}1.8235±2\pi k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}4.4597±2\pi k.$

Solve $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-0.2.$

$\theta \approx 1.7722±2\pi k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \approx 4.5110±2\pi k$

## Solving trigonometric equations in quadratic form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}u.\text{\hspace{0.17em}}$ If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

## Solving a trigonometric equation in quadratic form

Solve the equation exactly: $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta +3\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta -1=0,0\le \theta <2\pi .$

We begin by using substitution and replacing cos $\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =x.\text{\hspace{0.17em}}$ We have

${x}^{2}+3x-1=0$

The equation cannot be factored, so we will use the quadratic formula $\text{\hspace{0.17em}}x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.$

Replace $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ and solve. Thus,

Note that only the + sign is used. This is because we get an error when we solve $\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)\text{\hspace{0.17em}}$ on a calculator, since the domain of the inverse cosine function is $\text{\hspace{0.17em}}\left[-1,1\right].\text{\hspace{0.17em}}$ However, there is a second solution:

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

how to understand calculus?
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
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if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
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