The Student Room Group
Reply 1
Will probably get more help in maths :smile:
Reply 2
Use the identities Cos2x = Cos²x - Sin²x
and Cos²x + Sin²x =1
then solve the resultant quadratic for cosx
Reply 3
jjlang
I am looking for the workings for the following question.

Cos2X - 4CosX - 5 = 0


Use cos(2x) = 2cos²x - 1

cos(2x) - 4cosx - 5 = 0
-> 2cos²x - 1 - 4cosx - 5 = 0
-> 2cos²x - 4cosx - 6 = 0
-> 2[cos²x - 2cosx - 3] = 0
-> 2[(cosx - 3)(cosx + 1)] = 0

so cosx - 3 = 0, and cosx + 1 = 0
... cosx = 3 has no sol, cosx = -1, so x = arccos[-1] = π (180°)

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