TITLE: understanding intervals in trigonometry
QUESTION [5 upvotes]: So this is some I partially understand, I'm not sure what I don't and do understand because most of my understanding is based on assumptions... sorry if I sound a little stupid!
The equation $6 \cos x - \sin x = 5 $ needs to be turned into the form $ \Re  \cos (x + \alpha) $ then solved for in the interval $ -1/2\pi < x < 1/2\pi $.
I turned it into this: $ \sqrt{37} \cos( x + 0.165) $ then
 $ \sqrt{37} \cos( x + 0.165)  = 5$ 
$ \cos( x + 0.165)  = \frac{5}{\sqrt{37}}$
$ x + 0.165= \arccos( \frac{5}{\sqrt{37}} )$ 
$ x  = \arccos( \frac{5}{\sqrt{37}} ) - 0.165$ 
$ x = 0.44$ YAY! but... the answer is 0.44 and -0.771  
I'm thinking its asking what other value would of the above equation would end up in 5 also? Correct? How do I do this? Could someone explain what is meant by "solve this equation for that interval", and how does one go about it?
A problem I think might be related that I JUST cannot get my head around is this one:  

The angle made by a wasps wings horizontally is given by the equation $ \theta = 0.4 \sin600t $, where t is time is seconds. How many times a second does its wing oscillate? 

I tried solving this, honest but I do not know where to begin!

REPLY [2 votes]: As the cosine is periodic, there are many values of $\theta$ which have the same $\cos(\theta)$.  So they are just asking for all the values between $-\pi/2$ and $\pi/2$ that solve the equation.  Your solution got one of them, $\cos(0.605)$ does equal $5/\sqrt(37)$.  But so does $\cos(-.605)$  You were supposed to find that one, too. It leads to the solution -.771 when the .165 is subtracted.
For your second problem I presume the > sign is supposed to be *.  How much does the argument of the sine function (the thing you take the sine of) have to increase to go through one cycle?  How much does t have to increase to go through one cycle.  This gives you the period.  The frequency is one divided by this.