Question

Find all solutions x in the interval [0,2pi)
cos^2 (x/2) + 2cos (x/2) = -1

Answer 1

Have you considered simply factoring?

Can you solve this? \(x^{2} + 2x + 1 = 0\)?

Answer 2

Yes (x+1)^2=0

Answer 3

Excellent. Do EXACTLY the same thing with your original equation.

Answer 4

Hmm.. Would it be
cos(x/2)(cos(x/2)+2)=-1 ? Or am I wrong..

Answer 5

That was not EXACTLY the same thing.

Add 1 to both sides and give it another go. Remember, there is virtually no point to factoring unless one side of the equation is ZERO (0).

Answer 6

Okay.. so by adding 1 to each side, my new problem would be
cos^2(x/2)+2cos(x/2)+1=0 ..
and once factoring that, it would become
1/2 (4cos(x/2)+cos(x)+3) = 0
I still feel like that may be wrong..

Answer 7

I'm sorry. I'm not understanding this at all..

Answer 8

A few quick examples, starting with the one YOU already solved.

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

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

(2)
\(z^{2} + 2z + 1 = 0\)

\((z+1)^{2} = 0\)

(3)
\(Frog^{2} + 2Frog + 1 = 0\)

\((Frog+1)^{2} = 0\)

(4)
\(\sin^{2}(x) + 2\sin(x) + 1 = 0\)

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

(5)
\((Eiffel\;Tower)^{2} + 2(Eiffel\;Tower) + 1 = 0\)

\(((Eiffel\;Tower)+1)^{2} = 0\)

(6)
\(\cos^{2}(x/2) + 2\cos(x/2) + 1 = 0\)

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

EXACTLY like you did my little exercise. EXACTLY.

Answer 9

Oh! Okay, I see that. Didn't realize it would be that simple..
Now that is factored though, how do I find the solutions of x in the interval [0,2pi) ?

Answer 10

Again, EXACTLY like you would with the one that you solved.

\((x+1)^{2} = 0\) then \(x+1 = 0\;and\;x = -1\)

Let's see what you get.