Question

Please help

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0

Answer 1

u have
\[ \large \cos^2x+2\cos x+1=0 \]
set \(v=\cos x\) then u have
\[ \large v^2+2v+1=0 \]

Answer 2

cos 2x = cos (x+x) = cosxcosx-sinxsinx = cos^2 x - sin ^2 x

so your expression is cos^2 (x) - sin^2 (x) + 2 cos (x) + 1 = 0

sin^2 x is just 1- cos^2 (x)

to substitute: cos^2 (x) - (1-cos^2 (x)) + 2 cos (x) + 1 = 0
= cos^2 (x) - 1+ cos^2 (x) + 2 cos (x) + 1 = 0
= 2 cos^2 (x) + 2 cos (x) = 0
= 2 cos (x) (cos (x) + 1) = 0

so when cos x is 0 or cos x is -1, the whole expression will equal 0.

remember the unit circle. cos is 0 every pi/2 and every 3pi/2.
cos is -1 every pi.

so your values of x given the domain [0, 2pi] should be pi/2, 3pi/2 and pi.

Answer 3

Thanks so much man