Easy Q on interval change

[Sonal7]

The question states solve 3cos(x/2+45)=1 0<=x<=360 degrees. I am wondering how the interval is adjusted. I think the upper boundary changes to 360/2 +45=225, and the lower boundary is 45. Is this correct? I got the correct solution of 51 but just checking the boundaries are as i calculated.

 

[Dr.Peterson]

Why would you want to change the interval?

Probably this depends on the specific method you are using, so in order to help, we'll need to see your actual work.

I would solve for the general solution, and then find all solutions in the given interval. Presumably, you are doing something different.

 

[Sonal7]

I am querying the method. To solve for cos 2x =1 , 0<x<360 , we change the interval to 0<2x<540. That's what I originally meant to write. I shall post my workings in a few minutes when I get back to desk. I think in the original case one would divide 360 by 2 then add 45 degrees.

 

[pka]

The question states solve 3cos(x/2+45)=1 0<=x<=360 degrees. I am wondering how the interval is adjusted. I think the upper boundary changes to 360/2 +45=225, and the lower boundary is 45. Is this correct? I got the correct solution of 51 but just checking the boundaries are as i calculated.

In my mathematical-world view your question is:
\(3\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=1\)
\(\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{3}\)
\(x=2\left[\arccos\left(\frac{1}{3}\right)-\frac{\pi}{4}\right]\) SEE HERE
Now that you have a number solution, can you continue?

 

[Dr.Peterson]

I am querying the method. To solve for cos 2x =1 , 0<x<360 , we change the interval to 0<2x<540. That's what I originally meant to write. I shall post my workings in a few minutes when I get back to desk. I think in the original case one would divide 360 by 2 then add 45 degrees.

This is a very different approach than I use; so your question was not about the problem, but about your specific method, which you didn't state.

You never got back on this, so I'll comment on what you have said about it.

Using this method for the example here, the idea is not to find a new interval for x, but for the argument of the cosine. The interval 0<x<360 corresponds to 0<2x<720, by doubling each term. Why did you say 540? I may be missing part of what your method is, or you may just have typed wrong.

Now for the equation you were asking about,

3cos(x/2+45)=1 0<=x<=360 degrees

Here we want an interval containing x/2+45. So we can divide each term by 2, 0<=x/2<=360/2, and then add 45 degrees to each term:

0+45<=x/2+45<=360/2+45​

That is,

45<=x/2+45<=225​

So it appears that you did the right thing. You find all angles whose cosine is 1/3, between 45 and 225 degrees, and then solve for x in each case.

 

[Sonal7]

I didnt back to you as I was sick. I dont think it was coronavirus. I am still not 100% but back on the desk. I did a GCSE level type solving a trig equation. Nothing fancy at all. I think 51 degrees is only one in range.

 

[Dr.Peterson]

Yes, that's what I found.

My usual method (in part because that is what the textbook I most recently used for this does) is to find the general solution first:

3cos(x/2+45) = 1​

cos(x/2+45) = 1/3​

x/2+45 = 70.53+360k or -70.53+360k​

x/2 = 25.53+360k or -115.53+360k​

x = 51.06+720k or -231.06+720k​

The only solution in [0,360) is 51.06​

But your rejected solution disagrees with mine; where did you get 340.53 in your work?