All categories

3 years ago

Find all solutions in the region [0,2pi) for the equation cos^2(2x)-sin^2(2x)=0

1 answer:

4 0

$\cos^{2}(2x)-\sin^{2}(2x)=0$
$\rightarrow \sin^{2}(2x)=\cos^{2}(2x)$
$\rightarrow \frac{\sin^{2}(2x)}{\cos^{2}(2x)}=1$
$\rightarrow \tan^{2}(2x)=1$
$\rightarrow \tan(2x)=\pm 1$

$2x$ can range anywhere in $[0, 4\pi)$
So:
$\rightarrow 2x=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{13\pi}{4}, \frac{15\pi}{4}$
$\rightarrow x=\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8}$

You might be interested in

28 patients for 2 nurses

alukav5142 [94]

14 patients for each nurse

6 0

3 years ago

When flying at an altitude of 5 miles, the lines of sight to the horizon looking north and south make about a 173.7 degree angle

oksian1 [2.3K]

5 miles high is one of the sides of a triangle depending on accuracy level
h^2=x^2+y^2
we don't have 2 distances
Tan A=O/a
O=a tan A
We solve for O because the angle is at the top of the line going up and we want the opposite angle that is along the ground
O=5×tan(173.7/2)=90.854033512
The distance he can see is:
90.85*2~181.7 miles

Now we need to find the distance between lines:
The north south distance between each line is 69 miles
thus the number of degrees he will see will be:
181.7/69
=2 19/30

7 0

3 years ago

Deside whether g(x) = 2x^3 - 7x^2 - 3x^-1

mina [271]

You haven’t actually ask a questionnnnnn?

5 0

3 years ago

The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a

Leya [2.2K]

Answer:

a) 0.25

b) 52.76% probability that a person waits for less than 3 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \lambda e^{-\lambda x}$

In which $\lambda = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question:

$m = 4$

a. Find the value of λ.

$\lambda = \frac{1}{m} = \frac{1}{4} = 0.25$

b. What is the probability that a person waits for less than 3 minutes?

$P(X \leq 3) = 1 - e^{-0.25*3} = 0.5276$

52.76% probability that a person waits for less than 3 minutes

3 0

3 years ago

Need 2 more. 50 points!<br><br> Please show work, Thanks! :)

rjkz [21]

Answer:

1) C

2) C

Step-by-step explanation:

Question 1)

We want a parabola that has the vertex (-8, -7) and also passes through the point (-7, -4).

So, we can use the vertex form. Remember that the vertex form is:

$y=a(x-h)^2+k$

Where a is our leading co-efficient and (h, k) is our vertex.

We know that our vertex is (-8, -7). So, substitute this into the equation:

$y=a(x-(-8))^2+(-7)$

Simplify:

$y=a(x+8)^2-7$

Now, we need to determine the value of our a. To do so, we can use the point the problem had given us. We know that the graph passes through (-7, -4).

So, when x is -7, y is -4. Substitute -7 for x and -4 for y:

$(-4)=a((-7)+8)^2-7$