posted by Bryan .
If any body could help me figure out these problems. I would love to know how to do it, and not just get an answer, however if you don't have the time to explain it to me, then the answer will suffice.
6. (1 pt) Solve the following equation in the interval [0, 2 p
Note: Give the answer as a multiple of p. Do not use decimal
numbers. The answer should be a fraction or an integer. Note
that p is already included in the answer so you just have to enter
the appropriate multiple. E.g. if the answer is p/2 you should
enter 1/2. If there is more than one answer enter them separated
2(cos(t))2−cos(t)−1 = 0
t =________ pi
7. (1 pt) Solve the given equation in the interval [0,2 p].
Note: The answer must be written as a multiple of p. Give exact
answers. Do not use decimal numbers. The answer must
an integer or a fraction. Note that p is already provided with the
answer so you just have to find the appropriate multiple. E.g.
the answer is p
2 you should enter 1/2. If there is more than one
answer write them separated by commas.
2(sin x)2−5cos x+1 = 0
By the way when is says the interval, it's [0, 2pi] The p's should be pi's
For the first, factor the equation as you would if cos(t) were x.
For the second, recall that the #1 Identity (most important) is sin^2(x) + cos^2(x) = 1. Solve the identify for sin^2(x) and plug that into your equation so that it contains only cosine. Then, factor the equation as you would if cos(x) were simply x.
Try that, and if you have any problems, let us know.
So for the first the answers would be pi/3, pi, 5pi/3?
I tried it for the second problem and I couldn't make it work.
never mind...I got the first problem figured out. The second is still stumping me however.
Start by solving the #1 Identity: sin^2(x) + cos^2(x) = 1 for sin^2(x)...
sin^2(x) = 1 - cos^2(x)
Our equation is 2(sinx)^2 − 5cos(x) + 1 = 0
(sinx)^2 is the same as sin^2(x), so replace it with 1 - cos^2(x)...
2(1 - cos^2(x)) - 5cos(x) + 1 = 0
Distribute in the 2 and simplify. Then, factor.