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 .
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.
3 answers
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.
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
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 answer