Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1. So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi. For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have 2(tan^2 x + 1) + tan^2 x - 3 = 0 or 2tan^2 x + tan^2 x - 1 = 0 or 3 tan^2 x = 1. So x = pi/2, pi/2 + pi = 3pi/2.
hi the first thing I tried to solve this problem was to factor it so I have (cos x+1)(cos x+1) but now I dont know what to next
Unclear how to proceed. What is the first term, \(\cos(2x)\) or \(\cos^{2}(x)\)?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

cos \[\cos^{2}x\]
Perfect. The we have only to solve - by factoring. \(\cos^{2}x + 2\cos x + 1 = 0\) \((\cos(x) + 1)^{2} = 0\) Now what?
That is where I got stuck I am not sure what to do after factoring
What's the point of factoring? Why do we do that at all? If a*b = 0, what do we know about a or b? To contract, if a*b = 4, what do we NOT know about either a or b?
what each of the equal?
them*
Kind of. a*b = 0 tells us that either a = 0 or b = 0. What can we tell about a and b when it isn't zero? a*b = 4, for example? Is one of them zero? Aboslutely not, but that's about all we can tell. Is a = 4? Maybe. Is b = 12? Maybe. Not a clue. Only "=0" is particularly helpful. So, what should we do with that factored expression?
tkhunny your explanations are more complex than the problem itself :)
That's why we invent notation, so that problems can be simpler than explanations. If we have this: Factor1 * Factor2 = 0 Then we must have Either this Factor1 = 0 or this Factor2 = 0 How can we apply this to your factored equation?
cos x+1(0)=0 I am not sure
\((\cos(x)+1)^{2} = 0\) We have two factors, but hey are exactly the same. This limits the possible results. \(\cos(x) + 1 = 0\) \(\cos(x) = -1\) \(x = \dfrac{3\pi}{2} + 2k\pi\) where \(k\in \mathbb{Z}\) Do you know what all that means?
I do... but if cos(x)=-1 wouldn't [x=\pi\]
\[x=\]
x=pi*
anyway I got it ... thanks for the help!
Awesome. How did I do that?! Right you are. \(x = \pi + 2k\pi\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question