9.5 Solving trigonometric equations  (Page 9/10)

$\sqrt{3}{\cot }^{2}x+\cot x=1$

${\csc }^{2}x-3\csc x-4=0$

$20{\cos }^{2}x+21\cos x+1=0$

${\sec }^{2}x-2\sec x=15$

$\cos \left(-x\right)\sin x\cot x+{\sin }^{2}x$

$\sin (-x)\cos (-2x)-\sin (-x)\cos (-2x)$

$csc\left(\theta \right)\cot \left(\theta \right)\left({\sec }^2\theta -1\right)$

$${\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)\left(1+{\cot }^2\left(\theta \right)\right)\left(1+{\tan }^2\left(\theta \right)\right)$$

$\cos \left(\frac{7\pi }{12}\right)$

$\tan \left(\frac{3\pi }{8}\right)$

$\tan \left({\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\tan }^{-1}\sqrt{3}\right)$

$2\sin \left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{6}\right)$

$\cos \left(\frac{4\pi }{3}+\theta \right)$

$\tan \left(-\frac{\pi }{4}+\theta \right)$

${\cos }^2(\mathrm{32°}){\tan }^2(\mathrm{32°})$

$\cot \left(\frac{\theta }{2}\right)$

${\cos }^{2}x-{\sin }^{2}x-1=0$

${\cos }^{2}x=\cos x4{\sin }^{2}x+2\sin x-3=0$

$\cos \left(2x\right)+{\sin }^{2}x=0$

$2{\sin }^{2}x-\sin x=0$

Rewrite the expression as a product instead of a sum: $\cos \left(2x\right)+\cos \left(-8x\right).$

$8\cos \left(15x\right)\sin \left(3x\right)$

$2\left(\sin \left(8\theta \right)-\sin \left(4\theta \right)\right)$

Find all solutions of $\tan (x)-\sqrt{3}=0.$

Find the solutions of ${\sec }^{2}x-2\sec x=15$ on the interval $\left[0,2\pi \right)$ algebraically; then graph both sides of the equation to determine the answer.

$2\cos \left(\frac{\theta }{2}\right)=1$

$\sqrt{3}\cot \left(y\right)=1$

Find $\sin \left(2\theta \right),\cos \left(2\theta \right),$ and $\tan \left(2\theta \right)$ given $\cot \theta =-\frac{3}{4}$ and $\theta$ is on the interval $\left[\frac{\pi }{2},\pi \right].$

Find $\sin \left(\frac{\theta }{2}\right),\cos \left(\frac{\theta }{2}\right),$ and $\tan \left(\frac{\theta }{2}\right)$ given $\cos \theta =\frac{7}{25}$ and $\theta$ is in quadrant IV.

Rewrite the expression ${\sin }^{4}x$ with no powers greater than 1.

${\tan }^{3}x-\tan x{\sec }^{2}x=\tan \left(-x\right)$

$\sin \left(3x\right)-\cos x\sin \left(2x\right)={\cos }^{2}x\sin x-{\sin }^{3}x$

$\frac{\sin \left(2x\right)}{\sin x}-\frac{\cos \left(2x\right)}{\cos x}=\sec x$

Plot the points and find a function of the form $y=A\cos \left(Bx+C\right)+D$ that fits the given data.

The displacement $h(t)$ in centimeters of a mass suspended by a spring is modeled by the function $h(t)=\frac{1}{4}\sin (120\pi t),$ where $t$ is measured in seconds. Find the amplitude, period, and frequency of this displacement.

A woman is standing 300 feet away from a 2000-foot building. If she looks to the top of the building, at what angle above horizontal is she looking? A bored worker looks down at her from the 15 ^th floor (1500 feet above her). At what angle is he looking down at her? Round to the nearest tenth of a degree.

Two frequencies of sound are played on an instrument governed by the equation $n(t)=8\cos (20\pi t)\cos (1000\pi t).$ What are the period and frequency of the “fast” and “slow” oscillations? What is the amplitude?

The average monthly snowfall in a small village in the Himalayas is 6 inches, with the low of 1 inch occurring in July. Construct a function that models this behavior. During what period is there more than 10 inches of snowfall?

A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring $t$ seconds after being released. At what time will the spring come to rest? In this case, use 1 cm amplitude as rest.

Water levels near a glacier currently average 9 feet, varying seasonally by 2 inches above and below the average and reaching their highest point in January. Due to global warming, the glacier has begun melting faster than normal. Every year, the water levels rise by a steady 3 inches. Find a function modeling the depth of the water $t$ months from now. If the docks are 2 feet above current water levels, at what point will the water first rise above the docks?