7. Trig Inverses & Equations

United States of America

Grades 9-12

Lesson

When finding solutions to trigonometric inequalities we will see that the periodic nature of trigonometric functions often leads to more than one solution interval.

Find the solution for the following inequality:

$2\sin\left(4x+30^\circ\right)>\sqrt{3}$2`s``i``n`(4`x`+30°)>√3

**Think**: Let's first isolate the trigonometric function to determine if we can use the unit circle.

**Do**:

$2\sin\left(4x+30^\circ\right)$2sin(4x+30°) |
$>$> | $\sqrt{3}$√3 |

$\sin\left(4x+30^\circ\right)$sin(4x+30°) |
$>$> | $\frac{\sqrt{3}}{2}$√32 |

Notice the question is using degrees; therefore, make sure to use the appropriate angle measures. We know that $\sin60^\circ=\frac{\sqrt{3}}{2}$`s``i``n`60°=√32 on the unit circle in the first quadrant. Now we can create an inequality using the angle: $4x+30^\circ>60^\circ$4`x`+30°>60°.

$4x+30^\circ$4x+30° |
$>$> | $60^\circ$60° |

$4x$4x |
$>$> | $30^\circ$30° |

$x$x |
$>$> | $\frac{15}{2}^\circ$152° |

A trigonometric inequality must be solved, at least, within one common period. The common period of the $\sin$`s``i``n` function is the interval $\left[0^\circ,360^\circ\right]$[0°,360°]. To find the angles needed within the period, recall $\sin\theta=\sin\left(180^\circ-\theta\right)$`s``i``n``θ`=`s``i``n`(180°−`θ`), therefore $\sin60^\circ=\sin120^\circ$`s``i``n`60°=`s``i``n`120° .

$60^\circ$60° | $<$< | $4x+30^\circ$4x+30° |
$<$< | $120^\circ$120° |

$30^\circ$30° | $<$< | $4x$4x |
$<$< | $90^\circ$90° |

$7.5^\circ$7.5° | $<$< | $x$x |
$<$< | $22.5^\circ$22.5° |

**Reflect:** To turn the solution into general form, add $k$`k` times the common period. The common period for $\sin\theta$`s``i``n``θ` is $360^\circ$360°; however, the period was adjusted by the $4x$4`x` term within the angle $4x+30^\circ$4`x`+30°. This changes the period to:

$\frac{360^\circ}{4}=90^\circ$360°4=90°

So, we add $90^\circ k$90°`k` to each solution.

$7.5^\circ+90^\circ k$7.5°+90°k |
$<$< | $x$x |
$<$< | $22.5^\circ+90^\circ k$22.5°+90°k |

Find the solution for the following inequality:

$\cos(2x+\frac{\pi}{4})<\frac{1}{2}$`c``o``s`(2`x`+π4)<12

**Think**: The trigonometric function is already isolated. Now, determine if we can use the unit circle. Notice the question is using radians; therefore, make sure to use the appropriate angle measures.

**Do**: We know that $\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$`c``o``s`(π3)=12 on the unit circle in the first quadrant. A trigonometric inequality must be solved, at least, within one common period. The common period of the $\cos$`c``o``s` function is $[0,2\pi]$[0,2π]. To find the angles needed within the period, recall $\cos(\theta)=\cos(2\pi-\theta)$`c``o``s`(`θ`)=`c``o``s`(2π−`θ`) , therefore $$.

$\frac{\pi}{3}$π3 | $<$< | $2x+\frac{\pi}{4}$2x+π4 |
$<$< | $\frac{5\pi}{3}$5π3 |

$\frac{\pi}{12}$π12 | $<$< | $2x$2x |
$<$< | $\frac{17\pi}{12}$17π12 |

$\frac{\pi}{24}$π24 | $<$< | $x$x |
$<$< | $\frac{17\pi}{24}$17π24 |

**Reflect:** To turn the solution into general form, add $k$`k` times the common period. The common period for $\cos(\theta)$`c``o``s`(`θ`) is $2\pi$2π; however, the period was adjusted by the $2x$2`x` term within the angle $2x+\frac{\pi}{4}$2`x`+π4. This changes the period to:

$\frac{2\pi}{2}=\pi$2π2=π

So, we add$\pi k$π`k` to each solution.

$\frac{\pi}{24}+\pi k$π24+πk |
$<$< | $x$x |
$<$< | $\frac{17\pi}{24}+\pi k$17π24+πk |

Solve $\tan x>1$`t``a``n``x`>1.

Give the answer in interval notation, using $k$`k` as an arbitrary integer.

Solve $\sin\left(4x-60^\circ\right)\le0$`s``i``n`(4`x`−60°)≤0.

Give the answer in interval notation, using $k$`k` as an arbitrary integer.

Solve $\tan x>\frac{1}{\sqrt{3}}$`t``a``n``x`>1√3.

Give the answer in interval notation, using $k$`k` as an arbitrary integer.

Solve $\sin\left(5x-\frac{\pi}{3}\right)\le0$`s``i``n`(5`x`−π3)≤0.

Give the answer in interval notation, using $k$`k` as an arbitrary integer.

Solve $\sqrt{3}\sin x+\cos x>0$√3`s``i``n``x`+`c``o``s``x`>0.

Give the answer in interval notation, using $k$`k` as an arbitrary integer.

Some algebraic expressions involving trigonometric functions can be solved by hand, but for the most part reaching a solution will be very difficult, if not impossible.

Consider the inequality $\cos x\ge\frac{6}{13}$`c``o``s``x`≥613. What values of $x$`x` satisfy the expression? Using a graphing calculator, ($\cos^{-1}\left(\frac{6}{13}\right)$`c``o``s`−1(613)) find the value of $x$`x`, we can obtain the approximate solution

$x=62.514^\circ$`x`=62.514°

Therefore,

$360^\circ n-62.514^\circ\le x\le360^\circ n+62.514^\circ$360°`n`−62.514°≤`x`≤360°`n`+62.514°, for some integer $n$`n`, which can be written as $\left[360^\circ n-62.514^\circ,360^\circ n+62.514^\circ\right]$[360°`n`−62.514°,360°`n`+62.514°] in interval notation. This solution is not exact, but the accuracy is limited only by the precision of the computer system we are using. For most applications we can achieve as much precision as we like.

The key aspect of solving trigonometric inequalities graphically is to create two functions and to find the points of intersection of their graphs. From the example above, we can take $\cos x\ge\frac{6}{13}$`c``o``s``x`≥613 and look at the graphs of $f\left(x\right)=\cos x$`f`(`x`)=`c``o``s``x` and $g\left(x\right)=\frac{6}{13}$`g`(`x`)=613. We can then select an appropriate (or given) interval of values to see where $f\left(x\right)$`f`(`x`) intersects or is greater than $g\left(x\right)$`g`(`x`).

Use a graphing calculator to solve $8\cos\left(\frac{3x}{4}+58^\circ\right)\le\frac{41}{19}$8`c``o``s`(3`x`4+58°)≤4119 over the interval $\left(-360^\circ,360^\circ\right]$(−360°,360°]. Give your answer to three decimal places.

**Think**: There will be a general solution to this inequality, but we are only concerned with the values of $x$`x` that are greater than $-360^\circ$−360° and less than or equal to $360^\circ$360°. If we draw the graphs of $f\left(x\right)=8\cos\left(\frac{3x}{4}+58^\circ\right)$`f`(`x`)=8`c``o``s`(3`x`4+58°) and $g\left(x\right)=\frac{41}{19}$`g`(`x`)=4119, we can look at the points of intersection of the two graphs and see where $f\left(x\right)$`f`(`x`) lies below $g\left(x\right)$`g`(`x`).

**Do**: Here we have drawn each function over the interval $\left(-360^\circ,360^\circ\right]$(−360°,360°] using a computer graphing utility.

Using the intersection functionality, we can see that there are three points of intersection in the interval; one at $\left(-176.469^\circ,2.158\right)$(−176.469°,2.158), one at $\left(21.802^\circ,2.158\right)$(21.802°,2.158), and one at $\left(303.531^\circ,2.158\right)$(303.531°,2.158). From the image it is clear that $f\left(x\right)$`f`(`x`) is below $g\left(x\right)$`g`(`x`) between $x=-360^\circ$`x`=−360° and $x=-176.469^\circ$`x`=−176.469°, and between $x=21.802^\circ$`x`=21.802° and $x=303.531^\circ$`x`=303.531°. These two regions have been highlighted below.

We can write our solution set in interval notation as $\left(-360^\circ,-176.469^\circ\right]\cup\left[21.802^\circ,303.531^\circ\right]$(−360°,−176.469°]∪[21.802°,303.531°], where each value is rounded to three decimal places. This is read as "the union of the interval $\left(-360^\circ,-176.469^\circ\right]$(−360°,−176.469°] and the interval $\left[21.802^\circ,303.531^\circ\right]$[21.802°,303.531°]".

**Reflect**: Whether the boundaries of an interval in the solution set are open or closed depends on the type of inequality (whether we have $\le$≤ or just $<$<) and on the boundaries of the original interval under consideration. For example, $8\cos\left(\frac{3x}{4}+58^\circ\right)\le\frac{41}{19}$8`c``o``s`(3`x`4+58°)≤4119 is true when $x=-360^\circ$`x`=−360°, but $-360^\circ$−360° is not in the interval $\left(-360^\circ,360^\circ\right]$(−360°,360°], so $x=-360^\circ$`x`=−360° cannot be in the solution set. This is why we have designated an open lower boundary in the interval $\left(-360^\circ,-176.469^\circ\right]$(−360°,−176.469°].

Use a graphing calculator to solve $\cos\left(\frac{3x}{4}+1\right)\le\frac{4}{19}$`c``o``s`(3`x`4+1)≤419 over the interval $\left(-2\pi,2\pi\right]$(−2π,2π]. Give your answer to three decimal places.

**Think**: There will be a general solution to this inequality, but we are only concerned with the values of $x$`x` that are greater than $-2\pi$−2π and less than or equal to $2\pi$2π. If we draw the graphs of $f\left(x\right)=\cos\left(\frac{3x}{4}+1\right)$`f`(`x`)=`c``o``s`(3`x`4+1) and $g\left(x\right)=\frac{4}{19}$`g`(`x`)=419, we can look at the points of intersection of the two graphs and see where $f\left(x\right)$`f`(`x`) lies below $g\left(x\right)$`g`(`x`).

**Do**: Here we have drawn each function over the interval $\left(-2\pi,2\pi\right]$(−2π,2π] using a computer graphing utility.

Using the intersection functionality, we can see that there are three points of intersection in the interval; one at $\left(-3.145,0.211\right)$(−3.145,0.211), one at $\left(0.478,0.211\right)$(0.478,0.211), and one at $\left(5.233,0.211\right)$(5.233,0.211). From the image it is clear that $f\left(x\right)$`f`(`x`) is below $g\left(x\right)$`g`(`x`) between $x=-2\pi$`x`=−2π and $x=-3.145$`x`=−3.145, and between $x=0.478$`x`=0.478 and $x=5.233$`x`=5.233. These two regions have been highlighted below.

We can write our solution set in interval notation as $\left(-6.283,-3.145\right]\cup\left[0.478,5.233\right]$(−6.283,−3.145]∪[0.478,5.233], where each value is rounded to three decimal places. This is read as "the union of the interval $\left(-6.283,-3.145\right]$(−6.283,−3.145] and the interval $\left[0.478,5.233\right]$[0.478,5.233]".

**Reflect**: Whether the boundaries of an interval in the solution set are open or closed depends on the type of inequality (whether we have $\le$≤ or just $<$<) and on the boundaries of the original interval under consideration. For example, $\cos\left(\frac{3x}{4}+1\right)\le\frac{4}{19}$`c``o``s`(3`x`4+1)≤419 is true when $x=-2\pi$`x`=−2π, but $-2\pi$−2π is not in the interval $\left(-2\pi,2\pi\right]$(−2π,2π], so $x=-2\pi\approx-6.283$`x`=−2π≈−6.283 cannot be in the solution set. This is why we have designated an open lower boundary in the interval $\left(-6.283,-3.145\right]$(−6.283,−3.145].

Use a graphing calculator to solve $\sin x\le\frac{1}{2}$`s``i``n``x`≤12 over the interval $\left[0^\circ,360^\circ\right)$[0°,360°).

Use a graphing calculator to solve $\sec\left(\frac{x}{2}+50^\circ\right)<2.489$`s``e``c`(`x`2+50°)<2.489 over the interval $\left(-180^\circ,180^\circ\right]$(−180°,180°]. Give your answer to three decimal places.

Use a graphing calculator to solve $\cos^2\left(x\right)-\sin x<0.216$`c``o``s`2(`x`)−`s``i``n``x`<0.216 over the interval $\left[-90^\circ,270^\circ\right]$[−90°,270°]. Give your answer to three decimal places.

Use a graphing calculator to solve $2\cos x\ge\sqrt{3}$2`c``o``s``x`≥√3 over the interval $\left[0,2\pi\right)$[0,2π).

Use a graphing calculator to solve $3\csc x\le4.185$3`c``s``c``x`≤4.185 over the interval $\left(-\pi,\pi\right]$(−π,π].

Write your answer as a union of intervals, with each coordinate rounded to three decimal places.

Use a graphing calculator to solve $\cos^2\left(x\right)-\sin x<0.216$`c``o``s`2(`x`)−`s``i``n``x`<0.216 over the interval $\left[-\frac{\pi}{2},\frac{3\pi}{2}\right]$[−π2,3π2]. Give your answer to three decimal places.