Solve sec 2x = 2 between -Pi < x < Pi (can't find all the soln values)

Bronn

Junior Member
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Jan 13, 2017
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having trouble finding all the angles.

I can get:

sec 2x = 2
= cos 2x = 1/2

2x= pi/3
x= pi/6

positive cosine is quadrant (Q) 1 and 4

Q1= pi/6
Q4 = (2pi - pi/6)-2pi = -Pi/6 (adjusted to parameters)

theres two more angles however, 5Pi/6 and -5pi/6 which Im not sure how to get. The 2x I'm guessing means that its frequency is happening twice as much hence 4 angles. Just not sure how to get them. Ive tried messing around with not much luck.
any help?
cheers
 
having trouble finding all the angles.

I can get:

sec 2x = 2
= cos 2x = 1/2

2x= -(2π-π/3), -π/3, pi/3, (2π-π/3)
x = -5π/6, -π/6, π/6, 5π/6
x= pi/6

positive cosine is quadrant (Q) 1 and 4

Q1= pi/6
Q4 = (2pi - pi/6)-2pi = -Pi/6 (adjusted to parameters)

theres two more angles however, 5Pi/6 and -5pi/6 which Im not sure how to get. The 2x I'm guessing means that its frequency is happening twice as much hence 4 angles. Just not sure how to get them. Ive tried messing around with not much luck.
any help?
cheers
.
 
I'm sorry, but i need an explanation as I'm not understanding how to get the answer

2x= -(2π-π/3), -π/3, pi/3, (2π-π/3)

I dont know why its (2pi-pi/3)
 
1) cos(2x) = 1/2 has more than one solution, even on (-pi/2,pi/2). This is a natural result of the circle definition of the cosine. You must find them both.

2) cos(2x) = 1/2 has many more than one or two solutions. This is a natural result of the periodic nature of the cosine. If you found ALL the solutions suggested in the previous note, you can find all the others by adding or subtracting 2pi over and over and over... Generally, you can quit when you have enough.

Let's see what you get.
 
1) cos(2x) = 1/2 has more than one solution, even on (-pi/2,pi/2). This is a natural result of the circle definition of the cosine. You must find them both.

2) cos(2x) = 1/2 has many more than one or two solutions. This is a natural result of the periodic nature of the cosine. If you found ALL the solutions suggested in the previous note, you can find all the others by adding or subtracting 2pi over and over and over... Generally, you can quit when you have enough.

Let's see what you get.
but doesn't subtracting or adding 2pi push it outside the -pi<x<pi ?
 
but doesn't subtracting or adding 2pi push it outside the -pi<x<pi ?

The answer is, "maybe".

First, it behooves one to write ALL the solution and THEN pick the ones that are needed. No need to do the surgery up front.

For example, if I were soving \(\displaystyle sin(3x) = \sqrt{3}/2\), I might list \(\displaystyle 3x = \dfrac{\pi}{3} + 2k\pi\;and\;3x = \dfrac{2\pi}{3} + 2k\pi\), where k is an integer. That is infinitely many solutions.

Solving for x gives, \(\displaystyle x = \dfrac{\pi}{9} + \dfrac{2}{3}k\pi\;and\;x = \dfrac{2\pi}{9} + \dfrac{2}{3}k\pi\). These are the same infinitely many solutions.

If I also had a restriction, such as \(\displaystyle [0,2\pi]\), I would pick ALL values of k that contributed to that restriction. k = 0, obviously. That's two solutions. How about k = 1? Two more? Maybe k = -1? Perhaps k = 3? Keep looking until you find them ALL. In this example, we're moving only 2pi/3 at a time. That will NOT get us out of the desired range immediately.

In your problem, with 2x, we move a little faster. You also have negative values in your restricted range. Look both directions until you find them ALL.
 
edit* never mind i got it now. Thanks.
 
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