jfan17

\(2\cos^2{x} = \cos{x}\)

We can divide both sides by cosx

This gives us:

\(2\cos{x} = 1\)

Dividing by 2 on both sides, we get:

\(\cos{x} = \frac12\)

The values for which this is true in the interval \([0,2\pi)\) are 

\(\cos{\pi\over3}\) radians (cos 60 degrees) and also 

\(\cos{5\pi\over6}\)radians(cos 300 degrees)

Hey Asad. I think this error might just be occurring with you; it's most likely client-sided, and not server-sided as a whole. I posted an "off-topic" test post just a minute ago, and it was fine. Here's the link to it, and you can see for yourself:

https://web2.0calc.com/questions/test_16

I think it's better to not make assumptions so quickly for one or two posts that have run into moderation. That happened to me in the past too. Often times, there are keywords(according to melody) that are censored from the forum, which, when typed enough times, will cause a post to be flagged for moderation(regardless of category). Try revising your post and then seeing if it works once more. I don't think that this is any scenario to say "rip offtopic" :)

Ok, this needs to stop. I understand there may be 1 or 2 questions that may strike you as difficult, but when you post 10 questions in a single day, that should definitely raise some eyebrows. Are you posting on this forum to get the answers, or the solutions? Please try these questions yourselves before you post it, and explain what your thought process was and where your logic fell short when attempting these problems.

The number of ways of getting 7 are as follows:

1+6

2+5
3+4

4+3

5+2

6+1

that's 6 total cases, which means our answer is E). This is simply just a couple of cases. 

Correct me if I'm wrong, but is the answer not 15? Because the problem tells us that Paco has already won 15 tickets, and a bear costs 30 tickets, then he needs 30-15 = 15 more tickets to win that bear

If you look back to my soltuion, that allows me to add the two equations and cancel out all the odd terms. That's because when you substitute -1 for all the x values, the ones with all odd powers of x(x^1, x^3, etc.) will be negative, so when you add it with f(1), you get the even coefficients. Great question!

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

We know that we can't have a two digit factor in one of our three digit numbers(it can't fill in one digit), so we are left with:

1,2,3,4,6,8

Now, we can just bash out unique cases:

138

146

226

234

These are all the unique cases, for which you can reverse these to form any other arrangements that work(do you see the way I wrote it out systematically?) Notice that I write these numbers such that they're in ascending order, with the digits of each 3 digit number going in ascending order(in other words, the digit on the right of a given spot is either equal or greater in value than that digit). We now multiply out the number of ways we can arrange these to get our final answer.

138 = 3! = 6 ways

146 = 3! = 6 ways

226 = 3!/2! = 3 ways

234 = 3! = 6 ways

add all these up, and we get:

6+6+3+6 = 21 numbers, meaning our answer is E)

I noticed this was labeled practice, so I'm assuming its practice material for schoolwork. You should try this yourself first, before posting on this forum. If you already tried to solve the problem, then explain what you did, and where your logic falls short.

Here we make use of de moivre's theorem, which states that given a complex number z:

\(z=re^{i\theta} = r(\cos{\theta} + i\sin{\theta})\)

We can then raise z to the nth power to get a general form for this:

\(z^n = r^ne^{i*n*\theta} = r(\cos{n\theta} + i\sin{n\theta})\)

We can then set z as our desired complex number. We then get:

\(-3+3i\sqrt{3} = 6(\cos{2\pi\over3} + i\sin{2\pi\over3})\)

More generally, we can write:

\(-3+3i\sqrt{3} = 6(\cos(2n\pi +{2\pi\over3}) + i\sin(2n\pi + {2\pi\over3}))\)

\((-3+3i\sqrt{3})^{1/4} = 6(\cos(2n\pi/4 +{2\pi\over12}) + i\sin(2n\pi/4 + {2\pi\over12}))\)

Simplifying, we get:

\((-3+3i\sqrt{3})^{1/4} = 6(\cos(n\pi/2 +{\pi\over6}) + i\sin(n\pi/2 + {\pi\over6}))\)

Now we get our 4 roots from testing n from 0-3(it iterates after a cycle of 4). I'm sure you're capable enough to plug and chug in the numbers! :)

Edit: here's a good reference; it uses a similar method and explanation

https://socratic.org/questions/how-do-you-find-the-fourth-roots-of-i

Name cosx = y

4y² - 2y = 0

Dividing by 2 on both sides, we get:

2y² - y = 0

Dividing by y on both sides, we get:

2y -1 = 0

Adding 1 on both sides, weget:

2y = 1

y = 1/2

Substituting back in:

cosx = 1/2

the values for which this is true in the interval [0,2pi) are 

pi/6 and 5pi/6