# 1. Choose two strategies for solving the equation sec2x+8secX+12=0. Why do these strategies make the most scene?

2. Theta and a are each acute angles in standard position. sin=3/5 and cos=12/13. Explain why the direct angle measurements are not needed to find the compound trigonometric ratios.

3. Today the high tide in Matthews cove, New Brunswick, is at midnight. The water level at high tide is 11.5m. The depth, d in meters, of the water in the cove at the time t hours is modelled by the equation. d(t)=6+5.5cos(4t). Jenny is planning a day trip to the cove tomorrow, but the water needs to be at least 3m deep for her to manoeuvre her sailboat safely. How can Jenny determine the times it will when it will be safe for her to sail into Matthews cove? Explain your steps to solve it.

4. The tides at Cape Capstan change the depth of the water in the harbour. On one day in September, the tides have a high point of approximately 13m at 3 pm and 1.6m at 9 pm. The sailboat has a draft of 2.5m deep. The captain of the sailboat plans to exit the harbourer at 8:30 pm.

a. Write a cosine trigonometric equation that represents the situation

described.

b.Plot an accurate graph 1 cycle.

c.Determine whether the sailboat can exit the harbour safely.

## Looking at the pattern, I will assume you meant

sec^2 x + 8secx + 12 = 0

my first choice: let secx = t, then the equation becomes

t^2 + 8t + 12 = 0

(t + 2)(t + 6) = 0

t = -2 or t = -6

secx = -2 or secx = -6

cosx = -1/2 or cosx = -1/6

I assume you can proceed from here, you will get 2 answers

each for 0 ≤ x ≤ 2π

2nd choice:

sec2x+8secX+12=0

1/cos^2 x + 8/cosx + 12 = 0

multiply each term by cos^2 x

1 + 8cosx + 12cos^2 x = 0

(1 + 2cosx)(1 + 6cosx) = 0

cosx = -1/2 or cosx = -1/6 , reaching the same point as the first method

If you meant it the way your typed it

sec2x+8secx+12=0

1/cos 2x + 8/cosx + 12 = 0

1/(2cos^2 x - 1) + 8/cosx + 12 = 0

times cosx(2cos^2 x - 1)

cosx + 8(2cos^2 x -1) + 12cosx(2cos^2 x - 1) = 0

cosx + 16cos^2 x - 8 + 24cos^3 x - 12cosx = 0

let cosx = y

24y^3 + 16y^2 - 11y - 8 = 0

nasty cubic, Wolfram shows one real root, and two complexes, so

I doubt this is what you wanted.

2.

First of all, sin=3/5 and cos=12/13, are nonsense statements, you have left out the argument of sine , that is, you want something like sinθ = 3/5

Most compound trigonometric ratios have identity formulas

e.g. sin(A+B) = sinAcosB + cosAsinB

then for your data of sin a = 3/5 and cos b = 12/13, we can find

sin(a+b)

using standard right-angled triangles:

if sin a = 3/5, then cos a = 4/5, and if cos b = 12/13, then sin b = 5/13

sin(a+b) = (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65

3.

so you want

6+5.5cos(4t) ≥ 3

5.5cos(4t) ≥ -3

cos 4t ≥ -3/5.5

let's solve cos 4t ≥ -3/5.5 , and use the first answers

4t = cos^-1 (-3/5.5)

4t = 2.14773 or 4t = 4.13546

t = .5369 or t = 1.0338

Your equation cannot describe the movement of the tides anywhere.

The normal period of tidal movement is a bit more than 12 hours,

the period of your equation would be 2π/4 or appr 1.57 hours

so make the appropriate changes .

## The tides at Cape Capstan change the depth of the water in the harbour. On one day in September, the tides have a high point of approximately 13m at 3pm and 1.6m at 9pm. The sailboat has a draft of 2.5m deep. The captain of the sailboat plans to exit the harbourer at 8:30pm.

a. Write a cosine trigonometric equation that represents the situation

described.

b. Plot an accurate graph 1 cycle.

c. Determine whether the sailboat can exit the harbour safely.

## #4.

high tide = 13 m

low tide = 1.6

range = 11.4

so amplitude = 5.7 m

avg = (13+1.6)/2 = 7.3

period = 12 hours

for our k factor in the cosine function:

2π/k = 12

k = π/6

height(t) = 5.7 cos(π/6 t) + 7.3

right now we have a max of 13, when t = 0, .., 12, 18 , ...

we want our max to be at t = ... 15, ...

so we need to shift our graph to the right by 3 units

5.7 cos(π/6 (t - 3) ) + 7.3

check: when t = 15 , (3 pm),

height = 5.7cos(π/6(12)) + 7.3 = 13 , looks good

we want to know the height at 8:30 pm or t = 20.5

height = 5.7 cos(π/6 (20.5 - 3)) + 7.3

= 1.79 m

Since the draft of the boat is 2.5, it will hit bottom, stay put

## 1. The equation given is sec^2x + 8secx + 12 = 0. To solve this equation, we can use two strategies: factoring and the quadratic formula.

The factoring strategy involves trying to factor the equation into two binomial expressions. However, in this case, it is not possible to factor the equation directly.

Therefore, we can turn to the quadratic formula. The quadratic formula states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In our equation, a = 1, b = 8, and c = 12. Plugging these values into the quadratic formula, we get:

x = (-8 ± sqrt(8^2 - 4*1*12)) / (2*1)

Simplifying further, we have:

x = (-8 ± sqrt(64 - 48)) / 2

x = (-8 ± sqrt(16)) / 2

x = (-8 ± 4) / 2

This gives us two possible solutions:

x = (-8 + 4) / 2 = -2

x = (-8 - 4) / 2 = -6

The strategies of factoring and using the quadratic formula make the most sense in this case because they are general methods that can be applied to any quadratic equation. Additionally, factoring is often the preferred method when possible, as it can provide straightforward and simplified solutions.

2. When given the values of sin(theta) and cos(theta), we can use the Pythagorean identity to find the missing trigonometric ratios without needing to directly measure the angles.

The Pythagorean identity states that for any angle theta in standard position, sin^2(theta) + cos^2(theta) = 1.

Given sin = 3/5 and cos = 12/13, we can substitute these values into the Pythagorean identity:

(3/5)^2 + (12/13)^2 = 1

9/25 + 144/169 = 1

(9*169 + 144*25) / (25*169) = 1

After solving this equation, we can find the value of the missing trigonometric ratios, such as tan(theta), csc(theta), sec(theta), and cot(theta), by using the definitions of these ratios in terms of sin and cos.

The direct angle measurements are not needed because the trigonometric ratios are determined solely by the values of sin and cos, which can be obtained from the given information.