Lesson

Given a particular value of a trigonometric function, we may wish to find the values of the domain variable that are mapped to that function value. Or, given two distinct functions defined on the same domain, we may ask what values of the domain variable make the two functions equal.

These situations are what is meant by the idea of *solving *trigonometric (and other) equations. We are finding the values of the variable that make the equations true.

Given the function defined by $2\sin x$2`s``i``n``x` for $x\in[0,360^\circ]$`x`∈[0,360°], find the values of $x$`x` for which the function attains the value $1$1.

The problem is to solve the equation $2\sin x=1$2`s``i``n``x`=1, giving the solutions that are between $0$0 and $360^\circ$360°. The equation is rearranged to read $\sin x=\frac{1}{2}$`s``i``n``x`=12 and we proceed by asking what values of $x$`x` produce this function value.

From previous work we recall that $\sin30^\circ=\frac{1}{2}$`s``i``n`30°=12 and consequently that $\sin\left(180^\circ-30^\circ\right)=\sin150^\circ=\frac{1}{2}$`s``i``n`(180°−30°)=`s``i``n`150°=12. The sine function is negative between $\pi$π and $360^\circ$360°. So, these two are the only values that satisfy the equation.

We can give the solution as $x=30^\circ$`x`=30°, $150^\circ$150°.

If we had not known that $\sin30^\circ=\frac{1}{2}$`s``i``n`30°=12, it would have been necessary to use the inverse sine function. The steps are as follows:

$\sin x$sinx |
$=$= | $\frac{1}{2}$12 |

$x$x |
$=$= | $\arcsin\frac{1}{2}$arcsin12 |

$=$= | $30^\circ$30° |

A calculator gives only one solution since the inverse sine function is defined only on the interval $[-90^\circ,90^\circ]$[−90°,90°]. We subtract the calculator value from $180^\circ$180° to obtain the second solution, $180-30=150^\circ$180−30=150°.

Another way to consider the solution is to think of $2$2 functions $y=2\sin x$`y`=2`s``i``n``x` and $y=1$`y`=1. To solve $2\sin x=1$2`s``i``n``x`=1, is the same as graphically finding the intersection of these two separate lines.

You can see both solutions mentioned above here.

Solve $4\cos x+1=3$4`c``o``s``x`+1=3 for $x\in[0,360^\circ]$`x`∈[0,360°].

The equation is rearranged to $\cos x=\frac{1}{2}$`c``o``s``x`=12. In the domain $[0,360^\circ]$[0,360°], we can expect to find a solution somewhere in the interval $[0,90^\circ]$[0,90°] and another in the interval $[270^\circ,360^\circ]$[270°,360°] as these are the intervals in which the cosine function is positive.

Again, from previous work, we know that $\cos x=\frac{1}{2}$`c``o``s``x`=12 when $x=60^\circ$`x`=60°. The next solution must occur when $x=360-60$`x`=360−60. That is, when $x=300^\circ$`x`=300°.

Thus, the solutions are $x=60^\circ,300^\circ$`x`=60°,300°.

The graphs for this can also be used to find the solution. Graph both $y=4\cos x+1$`y`=4`c``o``s``x`+1 and $y=3$`y`=3 on the same set of axes. The intersections of these curves are the solutions.

Trigonometric equations do not always have easy solutions. It can be necessary to use a numerical method to obtain an approximate solution.

Find the measure in degrees of the angles satisfying $\sin\theta=\frac{1}{2}$`s``i``n``θ`=12 for $0^\circ\le\theta\le360^\circ$0°≤`θ`≤360°.

Consider the equation $\cos\theta=-0.7986$`c``o``s``θ`=−0.7986.

Find the measure in degrees of the acute angle satisfying $\cos\theta=0.7986$

`c``o``s``θ`=0.7986. Give your answer to the nearest degree.Find the measure in degrees of the angles satisfying $\cos\theta=-0.7986$

`c``o``s``θ`=−0.7986 for $0^\circ\le\theta\le360^\circ$0°≤`θ`≤360°. Give your answers to the nearest degree.

Find the measure in degrees of the angles satisfying $4\tan\theta+2=-2$4`t``a``n``θ`+2=−2 for $0^\circ\le\theta\le360^\circ$0°≤`θ`≤360°.

Form and use linear, quadratic, and simple trigonometric equations

Apply algebraic methods in solving problems