# Help with the Trigonometry Equation please!

• Feb 21st 2013, 08:09 AM
Danthemaths
Help with the Trigonometry Equation please!
Hi guys,
I am stressing myself out coz i dont know how to do this question.

Question being:
2cos(x-30)=1
0o≤X≤360o

I think you start by dividing both sides by 2 to get: cos(x-30)=1/2, but dont know what you do from here!

Dan

ps. could you please explain the working so I can understand it a bit better?

Thanks
• Feb 21st 2013, 08:22 AM
emakarov
Re: Help with the Trigonometry Equation please!
Look at the graph of cos(x) (note that it's not cos(x - 30)). For which values of x is it the case that cos(x) = 1/2? For each y is such that cos(y) = 1/2, we need to find x such that x - 30 = y. Find all such x's and then select those that are between 0 and 360.
• Feb 21st 2013, 08:26 AM
MINOANMAN
Re: Help with the Trigonometry Equation please!
Danthemaths

divide first by 2 to obtain cos(x-30)=1/2
which angle has cos =1/2 of course 60 therefore cos(X-30)=cos60
and of course the answer is x-30=60 which gives you x=90...study more the trigonometric equations...Tere are some other solutions negative but since you are restricted to 0....360 you cannot consider them.
• Feb 21st 2013, 08:35 AM
emakarov
Re: Help with the Trigonometry Equation please!
Quote:

Originally Posted by MINOANMAN
the answer is x-30=60 which gives you x=90... Tere are some other solutions negative but since you are restricted to 0....360 you cannot consider them.

There is another solution between 0 and 360.
• Feb 21st 2013, 08:43 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
Quote:

There is another solution between 0 and 360.
Thank you Minoanman!
and ema, is the other solution 360-90? which will give 570?

x=90
x=360-90=570

Is this correct?
• Feb 21st 2013, 08:55 AM
emakarov
Re: Help with the Trigonometry Equation please!
Quote:

Originally Posted by Danthemaths
is the other solution 360-90? which will give 570?

x=90
x=360-90=570

No, I suggest you follow the algorithm in post #2. And how do you get 570 from 360 - 90?
• Feb 21st 2013, 09:21 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
haha lol typo!

I meant 270 :P
• Feb 21st 2013, 09:46 AM
emakarov
Re: Help with the Trigonometry Equation please!
Quote:

Originally Posted by Danthemaths
I meant 270 :P

This is still not a solution of the original equation. You can check that 2cos(270 - 30) ≠ 1.
• Feb 21st 2013, 09:56 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
i still dont understand how u get second answer...?
• Feb 21st 2013, 09:59 AM
emakarov
Re: Help with the Trigonometry Equation please!
Quote:

Originally Posted by Danthemaths
i still dont understand how u get second answer...?

And I still don't understand what prevents you from following the steps in post #2. What exactly don't you understand about them?
• Feb 21st 2013, 10:11 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
ahh think this is right.

x=330?

If so, thanks alot guys :)
• Feb 21st 2013, 10:14 AM
emakarov
Re: Help with the Trigonometry Equation please!
Quote:

Originally Posted by Danthemaths
x=330?

Yes. We have cos(300) = cos(360 - 60) = cos(-60) = cos(60) = 1/2, so x - 30 = 300 implies that x = 330 is a solution.
• Feb 21st 2013, 10:15 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
thanks for helping me xD
• Feb 21st 2013, 10:25 AM
Danthemaths
Re: Help with the Trigonometry Equation please!
Quote:

Look at the graph of cos(x) (note that it's not cos(x - 30)). For which values of x is it the case that cos(x) = 1/2? For each y is such that cos(y) = 1/2, we need to find x such that x - 30 = y. Find all such x's and then select those that are between 0 and 360.
Also, on this website you used for the graph, it a pretty neat site. How did you get the x axis to appear as degrees instead of the default {0,1,2,3,4..}?