# understanding intervals in trigonometry

So this is some I partially understand, I'm not sure what I dont and do understand because most of my understanding is based on assumtions... 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

Im 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!

Thanks so much

-
+1 for thinking about it and showing what you have tried. – Ross Millikan Oct 25 '10 at 17:52

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.
@giddy: For the equation $\cos x=A$, the complete set of solutions is $x=\pm\arccos A + 2\pi n$ where $n$ is an arbitrary integer ($n=0,\pm 1,\pm 2,\ldots$). – Hans Lundmark Oct 25 '10 at 17:21
@giddy: (1) $\sin x=A$ if and only if $x=\arcsin A + 2\pi n$ or $x=\pi-\arcsin A + 2\pi n$. (2) $\tan x=A$ if and only if $x=\arctan A + \pi n$. – Hans Lundmark Oct 25 '10 at 17:30
If $\tan x=A$ the solutions are $\arctan A + \pi n$. If $\sin x=A$ the solutions are $\arcsin A + 2\pi n$ and $\pi - \arcsin A + 2\pi n$. You should be able to see this by drawing a horizontal line on a plot of each function. If you take a plot of $\cos x$, for example, draw a line at $y=.3$, say, and see where it intersects the graph. Those x values will be the solutions to $\cos x=.3$ and they follow the pattern Hans gave. – Ross Millikan Oct 25 '10 at 17:34
@giddy: Another way of seeing why it works is to look at the unit circle $x^2+y^2=1$. For the equation $\cos\phi=A$, draw the vertical line $x=A$ and see which angles $\phi$ that correspond to the points of intersection. For $\sin\phi=A$, do the same with the horizontal line $y=A$. And for $\tan\phi=A$, draw the line which passes through the origin and the point $(x,y)=(1,A)$. – Hans Lundmark Oct 25 '10 at 17:45