Question

Help Find all solutions in the interval [0, 2π).

cos2x + 2 cos x + 1=0

Answer 1

convert cos2x into cos^2x-sin^2

Answer 2

now reformulate it without sin

Answer 3

then use cos as X, then solve quadratic equation))

Answer 4

Okay so would the ansewr be \[x=\frac{ \pi }{ 2 }, \frac{ 3\pi }{ 2}\]?

Answer 5

is it \(\cos^2(x)\) or \(\cos(2x)\) for the first term?

Answer 6

cos^2(x)

Answer 7

then it is easier

Answer 8

\[\cos^2(x)+2\cos(x)+1=\left(\cos(x)+1\right)^2\]

Answer 9

therefore \(\cos(x)+1=0\iff \cos(x)=-1\) solve that one

Answer 10

x=2pi?

Answer 11

no

Answer 12

\(\cos(2\pi)=1\)