1. ## Trigonemetric Equation

cos t + 2 sec t = -3

(t = theta)

I need to determine all solutions of the given equation and express in radian measure.

So far I have:

cos^2 t + cos 0 + 2 = 0
(cost+1)(cost+2)

Then,
cost=-1 cost=-2

cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

Does this look correct?

2. Originally Posted by Snowcrash
cos t + 2 sec t = -3

(t = theta)

I need to determine all solutions of the given equation and express in radian measure.

So far I have:

cos^2 t + cos 0 + 2 = 0
(cost+1)(cost+2)

Then,
cost=-1 cost=-2

cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

Does this look correct?
Not quite.

$\displaystyle \cos(\theta) = -1 \, \, \therefore \, \, \theta = (2k+1) \pi \, \, \forall \, k \in Z$

In other words $\displaystyle \theta = \text{...} -5 \pi, \, -3 \pi, \, -\pi, \, \pi, \, 3\pi, \, 5\pi \text{...}$. All odd multiples of $\displaystyle \pi$.

3. Originally Posted by Snowcrash
cos t + 2 sec t = -3

(t = theta)

I need to determine all solutions of the given equation and express in radian measure.

So far I have:

cos^2 t + cos 0 + 2 = 0 don't understand this term
(cost+1)(cost+2)

Then,
cost=-1 cost=-2

cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

Does this look correct?
$\displaystyle \cos{t} + 2\sec{t} = -3$

$\displaystyle \cos^2{t} + 2 = -3\cos{t}$

$\displaystyle \cos^2{t} + 3\cos{t} + 2 = 0$

$\displaystyle (\cos{t} + 1)(\cos{t}+2) = 0$

$\displaystyle \cos{t} = -1$ ... $\displaystyle t = \pi$

$\displaystyle \cos{t} = -2$ ... no solution

4. I'm not sure where you got:

$\displaystyle \cos^{2}(x)+\cos(x)+2=0$

It should be:

$\displaystyle \cos^{2}(x)+3\cos(x)+2=0$

And you have factored it correctly in any case, but if you multiply it out, you will see that you do not get your original equation.

As for the next step:

$\displaystyle \cos(x)=-1$
$\displaystyle \cos(x)=-2$

We are being asked here to evaluate WHERE cosine is -1 and where cosine is -2. We know from the domain and range of cosine (and sine) that $\displaystyle \cos(x)=-2$ is a function with no solution. However, $\displaystyle \cos(x)=-1$ has the solution of $\displaystyle \pi$. Yet, is that the ONLY solution to the problem. Remember again the graph of sine and cosine: their domain is pos/neg infinity with a repeating graph yes? This would mean there are an INFINITE many number of solutions to our equation. We then need to write our answer in a manner that makes sense. Can you figure out how to write that solution?