Question

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0

Answer 1

When you have trigonometric functions with 'different-angles', try to bring them down to the same angle.\[cos 2x = 2cos^2 x - 1\]

So you can write your original equation as: \(2cos^2 x - 1 + 2 cos x + 1 = 0\)

Take \(cos x = y\)

And then solve the quadratic equation you get. :)

Answer 2

I don't get it, i thought we're suppose to factor?

Answer 3

Yes, you are supposed to factor!

After taking cos x = y
What do you think the equation becomes? :)

Answer 4

i think cos2x wann being here cos^2 x

or not ?

Answer 5

Yes

Answer 6

so is it (2cos+1)(cos-1) ? when factored ?

Answer 7

yes what ?

Answer 8

what i have supposed ?

Answer 9

yes it's suppose to be cos^2 x

Answer 10

cos^2 x +2cos x +1 =0 cos x = t

t^2 +2t +1 =0

(t+1)^2 =0

t+1=0

t = -1

cos x = -1

can you continue it ?

Answer 11

cos x= pi ?

Answer 12

not

cos x = -1 so then

x = arccos (-1)

do you know how many degree cosinus is equal -1 ?

Answer 13

180

Answer 14

\[1+\cos 2x+2\cos x=0\]
\[2 \cos ^2 x+2 \cos x=0\]
\[2 \cos x \left( \cos x+1 \right)=0\]
\[Either ~\cos x=0,~or~\cos x=-1\]
solve the two and get the solution.

Answer 15

\[\cos x=0=\cos \left( 2 n+1 \right)\frac{ \pi }{ 2 },n~ is~ any~ integer.\]
\[\cos x=-1=\cos \left( 2 n+1 \right)\pi,where ~n~is~any~integer.\]