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
by commas.
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
x=__________ pi
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.
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.
To solve these equations, you need to apply some mathematical techniques. I will explain the steps involved in solving each equation and guide you through the process.
1. Equation: 2(cos(t))^2 - cos(t) - 1 = 0
Interval: [0, 2π]
Step 1: Let's simplify the equation by substituting u = cos(t):
2u^2 - u - 1 = 0
Step 2: Solve this quadratic equation using factorization or the quadratic formula.
Factoring:
(2u + 1)(u - 1) = 0
Setting each factor equal to zero:
2u + 1 = 0 or u - 1 = 0
Step 3: Solve for u:
2u = -1 or u = 1
Step 4: Solve for t:
For u = -1/2:
cos(t) = -1/2
Considering the interval [0, 2π], the values of t where cos(t) = -1/2 are t = 2π/3 and t = 4π/3.
For u = 1:
cos(t) = 1
This corresponds to t = 0.
Therefore, the solutions in terms of π are: t = 0π, 2π/3, 4π/3
2. Equation: 2(sin(x))^2 - 5cos(x) + 1 = 0
Interval: [0, 2π]
Step 1: Let's simplify the equation by substituting u = sin(x):
2u^2 - 5√(1 - u^2) + 1 = 0
Step 2: Let's substitute another variable v = √(1 - u^2) to simplify further:
2(1 - v^2) - 5v + 1 = 0
2 - 2v^2 - 5v + 1 = 0
-2v^2 - 5v - 1 = 0
Step 3: Solve this quadratic equation using factorization or the quadratic formula.
Factoring:
(-v - 1)(2v + 1) = 0
Setting each factor equal to zero:
-v - 1 = 0 or 2v + 1 = 0
Step 4: Solve for v:
-v = 1 or 2v = -1
For v = 1:
√(1 - u^2) = 1
This corresponds to u = 0.
For v = -1/2:
√(1 - u^2) = -1/2
No real values of u satisfy this equation, so we discard it.
Step 5: Solve for x:
For u = 0:
sin(x) = 0
Considering the interval [0, 2π], the values of x where sin(x) = 0 are x = 0 and x = π.
Therefore, the solutions in terms of π are: x = 0π, π
By the way when is says the interval, it's [0, 2pi] The p's should be pi's
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.
what is this
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Solve the proportion. 16 /50 = x/ 156 . 25