Question

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

cos^2(x) + 2cos(x) + 1 = 0

Answer 1

I get cos(x)=-1
so the answers should be pi and 0 but the answer choices are:
x = 2π

x = π

x = pi divided by four., seven pi divided by four.

x = pi divided by two., three pi divided by two.

Answer 2

well, you're correct, \(\bf cos^{-1}(cos(x)) = cos^{-1}(-1)\)

Answer 3

\[\left( \cos x+1 \right)^{2}=0,\cos x=-1=\cos \pi \]
\[x=\pi \]

Answer 4

Since cos(x)=-1 the solution should have an x coordinate of -1, so shouldn't the solutions be pi, and 2pi

Answer 5

no cos is negative in second and third quadrants only

Answer 6

got it, thanks

Answer 7

yw