cos 2x - 4cosx - 5 = 0?

can you solve this ? and show me how please :)

Update:

thanks guys this really helped :)

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!

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