How can I find all the solutions of this?-VBForums

# Thread: How can I find all the solutions of this?

1. ## How can I find all the solutions of this?

Hey.
Okay, so I'm doing this question and have no idea how to find all the solution for cosΘ = 1/√2 when the domain is 0≤Θ≤2π (that's a pi symbol)
Here's what I have worked out so far.
cosΘ = 1/√2
= π/4
cos is positive in the 1st and 4th quadrants.

I know the second answer is 7π/4. How did the 7 get in the result? Is there an equation I have to do to get it?

Also, I am having trouble with cosΘ = 0 when the domain is the same as above.

Thank you.

2. ## Re: How can I find all the solutions of this?

Imagine the graph of the cosine function between 0 and 2*pi. Draw horizontal lines through the graph. The graph is symmetric about x=pi, so wherever the horizontal line intersects the graph, it must intersect the graph on the other side of x=pi as well. As an example, take the horizontal line y=1. In the domain 0 <= x <= pi, cos(x) = 1 only for x=0, so in the domain pi <= x <= 2pi, cos(x) = 1 only for x = pi + (pi - 0) = 2pi.

In your situation you should know (since it's one of the special angles) that cos(pi/4) = 1/sqrt(2) = sqrt(2)/2. Since 0 <= pi/4 <= pi, this solution can be reflected through x=pi to give a second solution, pi + (pi - pi/4) = 4pi/4 + 3pi/4 = 7pi/4.

More generally, suppose -1 < c <= 1. From the shape of the graph (in particular, the fact that it decreases on [0, pi] from 1 to -1 and mirrors itself after that), we have the following: there is a unique point where the horizontal line y=c intersects the graph cos(x) in the domain 0 <= x <= pi, i.e. there is a unique solution (call it t) to the equation cos(x) = c when we restrict x to the range 0 <= x <= pi. Moreover, there is a corresponding unique solution t' to this equation in the range pi <= x <= 2pi. The two solutions are related by the fact that the distance from pi to t is the same as the distance from t' to pi, since the graph mirrors about x=pi. Algebraically, pi - t = t' - pi.

Given a solution t, we can compute t' since t' = pi + (pi - t). For c = 1/sqrt(2), the above holds, and we know that t=pi/4, so t'=pi + (pi - pi/4) = 7pi/4 as above. Moreover these are the only solutions on [0, 2pi], since t is unique on [0, pi] and t' is unique on [pi, 2pi].

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