Answer to Question #173523 in Trigonometry for Jon jay Mendoza

Question

Find the solution of the equation in the interval 0⁰ ≤ 𝑥 < 360⁰ 

1. cos x + √𝟑 = - cos x
2. sin² x – tan x cos x = 0 .
3. sin x + √𝟐 = − sin x 
4. 2 cos² x – 5 cos x = 3
5. √𝟑 csc x + 2 = 0

Solve the worded problem:

Example 6-https://ibb.co/41tVQZt

1. Look back at the model in Example 6 on page 7. On which days of the year are there 10 hours of sunlight in Prescott, Arizona?

2. The tide, or depth of the ocean near the shore, changes throughout the day. The depth of the Bay of Fundy can be modeled by:

link model-https://ibb.co/hmrm7GN

where d is the water depth in feet and t is the time in hours. Consider a day in which t = 0 represents 12:00 A.M. At what time(s) is the water depth 3 1/2 feet

Answer 1

1.

$$\cos x+\sqrt{3}=-\cos x$$

$$\cos x=-\dfrac{\sqrt{3}}{2}, 0\degree\leq x<360\degree$$

$$x=150\degree\ \text{or} \ x=210\degree$$

2.

$$\sin^2 x-\tan x \cos x=0$$

$$\sin^2 x-\dfrac{\sin x}{\cos x}\cdot\cos x=0, \cos x\not=0$$

$$\sin x(\sin x-1)=0, 0\degree\leq x<360\degree$$

$$x=0\degree \text{or}\ x=180\degree$$

3.

$$\sin x+\sqrt{2}=-\sin x$$

$$\sin x=-\dfrac{\sqrt{2}}{2}, 0\degree\leq x<360\degree$$

$$x=225\degree\ \text{or} \ x=315\degree$$

4.

$$2\cos^2 x-5\cos x=3$$

$$2\cos^2 x-5\cos x-3=0$$

$$\cos x=-\dfrac{1}{2} \ \text{or} \ \cos x=3, -1\leq \cos x\leq 1$$

Then

$$\cos x=-\dfrac{1}{2}, 0\degree\leq x<360\degree$$

$$x=120\degree \text{or}\ x=240\degree$$

5.

$$\sqrt{3}\csc x+2=0$$

$$\sin x=-\dfrac{\sqrt{3}}{2}, 0\degree\leq x<360\degree$$

$$x=240\degree\ \text{or} \ x=300\degree$$

1.

$$10=2.325\sin {\dfrac{\pi}{6}(t-2.667)}+12.155$$

$$2.325\sin {\dfrac{\pi}{6}(t-2.667)}=10-12.155$$

$$\sin {\dfrac{\pi}{6}(t-2.667)}=-\dfrac{2.155}{2.325}$$

$$\dfrac{\pi}{6}(t-2.667)\approx(-1)^n\sin^{-1}(-0.927)+\pi n, n\in\Z$$

$$\dfrac{\pi}{6}(t-2.667)\approx-1.186$$

$$t\approx0.402$$

$t_1=12\text{ days}$

Or

$$\dfrac{\pi}{6}(t-2.667)\approx\pi-1.186$$

$$t\approx10.932$$

$t_2=10\text{\ months plus 28 days}$

2.

$$d=35-28\cos {(\dfrac{\pi}{6.2}t})$$

$$3.5=35-28\cos {(\dfrac{\pi}{6.2}t})$$

$$\cos {(\dfrac{\pi}{6.2}t})=1.125$$

No solution