# finding solutions for cos

• November 12th 2010, 03:04 PM
finding solutions for cos
I have to find the two solutions for;

$
cos(2t)=-0.75, \pi < t < 2 \pi
$

i have sketched a graph, but am unsure how to get the values mathematically?

Many thanks :)
• November 12th 2010, 03:29 PM
skeeter
Quote:

I have to find the two solutions for;

$
cos(2t)=-0.75, \pi < t < 2 \pi
$

$\pi < t < 2\pi$

$2\pi < 2t < 4\pi$

$\cos(2t) = -0.75$

$2t = 2\pi + \arccos(-0.75)$

$t = \pi + \frac{1}{2}\arccos(-0.75)$

$2t = 4\pi - \arccos(-0.75)$

$t = 2\pi - \frac{1}{2}\arccos(-0.75)$
• November 12th 2010, 03:46 PM
the question then goes on to say, "if the solutions are denoted by t1 and t2 where t1 < t2, find t1 and t2?"

• November 12th 2010, 03:50 PM
skeeter
Quote:

the question then goes on to say, "if the solutions are denoted by t1 and t2 where t1 < t2, find t1 and t2?"

you can't tell which solution for t is the largest or smallest?
• November 12th 2010, 03:56 PM
• November 12th 2010, 04:15 PM
Quote:

Originally Posted by skeeter
you can't tell which solution for t is the largest or smallest?

i can, but i need to know which range, both seems to satisfy the t1 < t2 criteria??? but which is which, if you see what i mean?
• November 12th 2010, 04:17 PM
skeeter
one more time ...

$t = \pi + \frac{1}{2}\arccos(-0.75)$

$t = 2\pi - \frac{1}{2}\arccos(-0.75)$

evaluate each value of t in your calculator ... check the results with your graph.
• November 12th 2010, 04:27 PM
so because the summary of the statement is a just a $t$ as apposed to $2t$, then the ones with $t$ are the real answers? would this statement be correct?
so because the summary of the statement is a just a $t$ as apposed to $2t$, then the ones with $t$ are the real answers? would this statement be correct?