1. ## [SOLVED] trig problem

cosY= -sqrt3/2

and

cosY= -sqrt3/2 for y in [pi,2pi]

The question ask to solve these equations

2. Originally Posted by john doe
cosY= -sqrt3/2

and

cosY= -sqrt3/2 for y in [pi,2pi]

Keep in mind that cosine is negative in the second and third quadrants.

The second question is asking for particular angles within the interval $\pi\leq y\leq2\pi$. Note that you will only have one solution.

keep in mind that $\cos\left(\tfrac{\pi}{6}\right)=\tfrac{\sqrt{3}}{2 }$ Change the angle to get $\cos(?)=-\tfrac{\sqrt{3}}{2}$

The first question wants you to generalize the solution. This can be a little tricky. Find several solutions to the equation, and then try to find a pattern given in the coefficients of the angles. Try this one and let us know where you get stuck...if you get stuck...

I hope this helps.

--Chris

3. ## Check this out

Originally Posted by john doe
cosY= -sqrt3/2

and

cosY= -sqrt3/2 for y in [pi,2pi]