Question

How do you solve `cos 2x= cos x-1`?

Answer 1

`x = 60^@ , 90^@ , 270^@ and 300^@`

Explanation:

Using trig formulae:

`cos 2x = cos^2x - sin^2x = cos^2x - ( 1 - cos^2x ) = 2cos^2x - 1`

Replace cos2x by `(2cos^2 x - 1 )`

cos2x = cox - 1 becomes `2cos^2x - 1 = cosx - 1`

This is a quadratic function and to solve equate to zero.

hence : `2cos^2x - 1 - cosx + 1 = 0`

simplifies to : `2cos^2x - cosx = 0`

factorise : cosx (2cosx - 1 ) = 0

→ cosx = 0 → x = `90^@ , 270^@`

and cosx `= 1/2 →x = 60^@ , 300^@`

These solutions are in the interval 0 < x ≤ 360