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!
Best Answer
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.