Question

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0

Answer 1

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.

Answer 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

Answer 3

Unclear how to proceed. What is the first term, \(\cos(2x)\) or \(\cos^{2}(x)\)?

Answer 4

cos \[\cos^{2}x\]

Answer 5

Perfect. The we have only to solve - by factoring.

\(\cos^{2}x + 2\cos x + 1 = 0\)

\((\cos(x) + 1)^{2} = 0\)

Now what?

Answer 6

That is where I got stuck I am not sure what to do after factoring

Answer 7

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?

Answer 8

what each of the equal?

Answer 9

them*

Answer 10

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?

Answer 11

tkhunny your explanations are more complex than the problem itself :)

Answer 12

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?

Answer 13

cos x+1(0)=0 I am not sure

Answer 14

\((\cos(x)+1)^{2} = 0\)

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?

Answer 15

I do... but if cos(x)=-1 wouldn't [x=\pi\]

Answer 16

\[x=\]

Answer 17

x=pi*

Answer 18

anyway I got it ... thanks for the help!

Answer 19

Awesome. How did I do that?! Right you are. \(x = \pi + 2k\pi\)