can you solve this ? and show me how please
thanks guys this really helped
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
cos^2 x+4 cos x-5=0
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.
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
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!
{x= -Pi}, {x = Pi}
Comments
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
cos^2 x+4 cos x-5=0
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
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.
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
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!
{x= -Pi}, {x = Pi}