Trignometry help!?

Question

find all solutions in the range 0< x < (PI)

cos2x - cos4x =sin x

giving all the solutions in (PI) radians

Im really stuk on this Q...if ny one can solve this plz could u explain it aswell..thanks.

Answer 1

You can use
cos(2x) = 1 - 2 sen(x)^2 and sen(2x) = 2 sen(x) cos(x)

cos(2x) - cos(4x) = 1 - 2 sen(x)^2 - (1 - 2 sen(2x)^2) = - 2 sen(x)^2 - 8 sen(x)^2 cos(x)^2 = -2 sen(x)^2 (1+4cos(x)^2) = 2 sen(x)^2 (4 sen(x)^2 - 3)

then 2 sen(x)^2 (4 sen(x)^2 - 3) = sin(x)

a general metod to solve this is to assume sen(x) = t

t (t (8 t^2 -6 ) - 1) = t (8 t^3 - 6 t -1) = 0

t = 0 (<=> x=0,PI not valid)

and look for t solutions of 8 t^3 - 6 t - 1 = 0 where 0
Unlucky is a third grade equation.. but We know that t = sin(x) and we can add the knowdlegde about sin(x): (Yes in some way has been an useless assumption.. in this case..)

8 sin(x)^3 - 6 sin(x) -1 = 8 sin(x) (1-cos(x)^2) - 6 sin(x) - 1 = 2 sin(x) (1-4cos(x)^2) = 2 sin(x) (1 + 2 cos(x)) (1 - 2 cos(x)) = 0

Then valid solution are
1 + 2 cos(x) = 0
1 - 2 cos(x) = 0

Ok?

Bye
Enrico