cos4x-7cos2x=8?

Question

how would you solve for x?

Answer 1

cos(4x) = cos(2(2x)) = 2cos²(2x) – 1

cos(4x) – 7cos(2x) = 8
(2cos²(2x) – 1) – 7cos(2x) = 8
2cos²(2x) – 7cos(2x) – 9= 0
2(cos²(2x) – 2(7/4)cos(2x) + 49/16 – 49/16) – 9= 0
2(cos(2x) – 7/4)² – 49/8 – 9 = 0
2(cos(2x) – 7/4)² = 121/8
(cos(2x) – 7/4)² = 121/16
cos(2x) – 7/4 = ± 11/4
cos(2x) = (7± 11)/4

cos(2x) = 9/2 ó cos(2x) = –1

Observe that –1 ≤ cos(x) ≤ 1 for all x in R, then

cos(2x) = –1

=> 2x = (2n+1)π , with n in Z (Integers) (Because Cos reaches –1 in all the odd multiples of π)

x = n π +½ π with n in Z

Answer 2

Use cos2y = 2cos^2 y -1 and write cos4x= 2cos^2(2x)

Thus alfter some algebra you have 2cos^2(2x) -7cos2x -9=0 which is a quadratic equation in cos2x. Solve then for cos2x = root1 or root2 and the rest is easy!

Answer 3

I'd write cos 4x in terms of cos 2x.

I'd then have an easy quadratic equation for cos2x.

I'd solve it and see where I was.

Any solution for cos 2x with absolute value > 1 would of course be immediately discarded.