`cos (2x) = -5cos x - 4`

Using the double angle identity of cosine, which is `cos 2u =` `2cos^2u - 1` , we will have:

`2cos^2x - 1 = -5cos x - 4`

`2cos^2x - 1 + 5cosx + 4 = 0`

`2cos^2x + 5cos x + 3 = 0`

Then, treat the above expression as a quadratic equation. And factor.

`(2cos x + 3) ( cosx + 1) = 0`

Set each factor to zero.

`2cos x - 3 = 0 ` and `cos x + 1 = 0`

`2cos x = 3` `cos x = -1`

`cos x = 3/2`

Note that the range of a cosine function is from -1 to 1 only. So, we will not cosider cos x = 3/2 as our solution.

Hence,

` x = pi`

And the general solution is:

` x = pi + 2pik`

**Answer:**`x = pi + 2pik`

cos2x = 2cos^2(x)-1

cos(2x)=-5 cosx-4

2cos^2(x)-1 = -5 cosx-4

2cos^2(x)-1+5 cosx+4 = 0

2cos^2(x)+5 cosx+3 = 0

Let cosx = t

2t^2+5t+3 = 0

Then t=(-5+sqrt(5^2-4*2*3))/(2*2)

=(-5+sqrt(25-24))/(4)

= (-5+1)/4

= -1

The next answer;

t = (-5-sqrt(5^2-4*2*3))/(2*2)

=(-5-sqrt(25-24))/(4)

= (-5-1)/4

= -1.5

If t=-1 then;

cosx=-1

cosx=cos (pi)

= **2*n*pi+pi ** (n is an integer)

If t=-1.5 then;

cosx = -1.5

This cannot be happen because -1<=cosx<=1

So t=-1.5 is not a answer.

cos (2x) = -5 cos x - 4

Since cos (2x) = 2 cos x^2 - 1

2 cos x^2 - 1 = -5 cos x - 4

2 cos x^2 + 5cos x + 3 = 0

( 2 cos x + 3 ) ( cos x + 1 ) = 0

cos x = - 3 / 2 or cos x = - 1

Since cos x <= 1 , cos x not = -3/2

cos x = - 1

x = cos ^ - 1 ( - 1)

x = pi +/- 2 n pi , where n = 1, 2, 3 ...