cos 2x - 4cosx - 5 = 0?

vicky-edwards_60c0e1da74f2c

can you solve this ? and show me how please

Update:

thanks guys this really helped

ed-i

If you really mean cos 2x, and not cos^2 x, then:

cos 2x - 4 cos x - 5 = 0

(2 cos^2 x - 1) - 4 cos x - 5 = 0

2 cos^2 x - 4 cos x - 6 = 0

cos^2 x - 2 cos x - 3 = 0

(cos x - 3)(cos x + 1) = 0

cos x - 3 = 0 or cos x + 1 = 0

cos x = 3 (reject).....cos x = -1

................................x = π

trish-e

cos^2 x+4 cos x-5=0

anonymous

cos 2x - 4 cos x - 5 = 0

(2 cos^2 x - 1) - 4 cos x - 5 = 0

2 cos^2 x - 4 cos x - 6 = 0

cos^2 x - 2 cos x - 3 = 0

(cos x - 3)(cos x + 1) = 0

cos x - 3 = 0 or cos x + 1 = 0

cos x = 3 (reject).....cos x = -1

kevin-melkowski

cos2x = 2cos^2x-1 per double angle theorem.

So 2cos^2x -1 -4cos x -5 = 0

Now substitute u = cos x.

2u^2 - 4u - 6 <--- this looks familiar.

2u^2 + 2u - 6u - 6

2u (u + 1) - 6 (u+1)

So (2u - 6) (u + 1) so u = 3 and -1. Now substitute cos x back in for u.

cos x = 3 and -1, but since you can't have anything above or below 1 and -1 you can't have 3.

So cos x = -1 which is also pi.

gile

cos2x - 4cosx - 5 = 0

2cos²x - 1 - 4cosx - 5 = 0

2cos²x - 4cosx - 6 = 0

cos²x - 2cosx - 3 = 0

(cosx + 1)(cosx - 3) = 0

cosx - 3 ≠ 0

Hence

cosx + 1 = 0

cosx = -1

x = (2k + 1)π

anonymous

By the cosine double-angle formula:

cos(2x) = 2cos^2(x) - 1.

Then:

cos(2x) - 4cos(x) - 5 = 0

==> [2cos^2(x) - 1] - 4cos(x) - 5 = 0

==> 2cos^2(x) - 4cos(x) - 6 = 0

==> 2[cos^2(x) - 2cos(x) - 3] = 0

==> 2[cos(x) - 3][cos(x) + 1] = 0

==> cos(x) = -1 and cos(x) = 3.

However, since -1 <= cos(x) <= 1, this equation is satisfied when:

cos(x) = -1.

This occurs when x = π ± 2πk, where k is an integer.

I hope this helps!

katie-evans_60c0e1deead72

{x= -Pi}, {x = Pi}