# How can I solve an equation based off of a quadrant and equation form given an angle?

Given that 3pi/2 < z < 2pi

x = arccos(sin(z))

Given different values for z (which are angles on the unit circle) how would I write the results in these two forms, where C is a constant?:

a.) x = z + C

b.) x = -z + C

Say angle z, = 7pi/4 and z = 5pi/3

I've drawn the unit circle and everything..

For example, when z = 7pi/4, I know x would = arccos(sin(z)).. z = arccos(-sqrt(2)/2)

I also know that cos = (sqrt(2)/2) at 5pi/4 and 3pi/4

But I'm not sure how to represent those in one of the forms given.

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Let's do the first one, where $z=\frac{7\pi}{4}$. The angle $z$ is in the fourth quadrant. Its sine is $-\frac{1}{\sqrt{2}}$. Now take the $\arccos$ of this number. since the number is negative, the $\arccos$ of the number, that is, $x$, is $\frac{3\pi}{4}$.

Now we are asked to find a number $C$ such that $x=z+C$. Since $x=\frac{3\pi}{4}$ and $z=\frac{7\pi}{4}$, we get $C=\frac{3\pi}{4}-\frac{7\pi}{4}$. So $C=-\pi$.

The second question asks for $x=-z+C$. So $C=\frac{3\pi}{4}+\frac{7\pi}{4}=\frac{5\pi}{2}$.

The other problrm can be done using similar reasoning.

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Thanks so much! Made perfect sense. –  ModdedLife Apr 22 '13 at 7:37
Which one would I use though? –  ModdedLife Apr 22 '13 at 7:42
@ModdedLife: Questions a) and b) were two separate questions, and they were both answered above. In a practical situation, either one would do the job, with a) maybe somewhat more useful. –  André Nicolas Apr 22 '13 at 7:47
By the way, for the question about $\frac{5\pi}{3}$, that's $\frac{\pi}{3}$ ($60$ degrees) short of $2\pi$. The sine is $-\frac{\sqrt{3}}{2}$. The $\arcos$ of this is $60$ degrees past $90$, so $\frac{5\pi}{6}$. –  André Nicolas Apr 22 '13 at 7:55