# Thread: [SOLVED] trig problem

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]

The question ask to solve
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]

The question ask to solve
cosY= -sqrt3/2 for y in [pi,2pi]
cosY is negative in third quadrant that is in interval [pi,3pi/2]
if cosY=sqrt3/2 then
Y=pi/6 (taking principle value)
so value of Y for which cosY=-sqrt3/2 will be
=pi+pi/6=7pi/6 (for third quadrant)
therfor y=7pi/6
hope this helps