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

0^{o}≤X≤360^{o}

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!

Thanks in advance

Dan

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

Thanks

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.

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.

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.

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?

So the answers will be

x=90

x=360-90=570

Is this correct?

Re: Help with the Trigonometry Equation please!

Quote:

Originally Posted by

**Danthemaths** is the other solution 360-90? which will give 570?

So the answers will be

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?

Re: Help with the Trigonometry Equation please!

haha lol typo!

I meant 270 :P

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.

Re: Help with the Trigonometry Equation please!

i still dont understand how u get second answer...?

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?

Re: Help with the Trigonometry Equation please!

ahh think this is right.

x=330?

If so, thanks alot guys :)

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.

Re: Help with the Trigonometry Equation please!

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..}?