I have to find the two solutions for;

$\displaystyle

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

$

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

Many thanks :)

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- Nov 12th 2010, 03:04 PMmadmax29finding solutions for cos
I have to find the two solutions for;

$\displaystyle

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

$

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

Many thanks :) - Nov 12th 2010, 03:29 PMskeeter
$\displaystyle \pi < t < 2\pi$

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

$\displaystyle \cos(2t) = -0.75$

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

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

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

$\displaystyle t = 2\pi - \frac{1}{2}\arccos(-0.75)$ - Nov 12th 2010, 03:46 PMmadmax29
the question then goes on to say, "if the solutions are denoted by t1 and t2 where t1 < t2, find t1 and t2?"

what are the answers? - Nov 12th 2010, 03:50 PMskeeter
- Nov 12th 2010, 03:56 PMmadmax29
is it 8.70 rads & 10.15 rads, or 4.35 rads & 5.07 rads????

- Nov 12th 2010, 04:15 PMmadmax29
- Nov 12th 2010, 04:17 PMskeeter
one more time ...

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

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

evaluate each value of t in your calculator ... check the results with your graph. - Nov 12th 2010, 04:27 PMmadmax29
so because the summary of the statement is a just a $\displaystyle t$ as apposed to $\displaystyle 2t$, then the ones with $\displaystyle t$ are the real answers? would this statement be correct?

- Nov 12th 2010, 05:05 PMmr fantastic