Question

How do you solve `cos^2x-sin^2x= -cosx`?

Answer 1

It is

#cos^2x-sin^2x= -cosx=>cos^2x-(1-cos^2x)=-cosx=> 2*cos^2x+cosx-1=0=>(cosx+1)*(2*cosx-1)=0#

Hence form the last equation we get

`cosx=-1=>x=2*k*pi+-pi` where `k` is an integer

`2*cosx-1=0=>cosx=1/2=>x=2*n*pi+-pi/3` where `n` is an integer.

Finally the solutions are

`(2*k*pi+-pi,2*n*pi+-pi/3)`

Answer 2

Final solution set

`x = {0^@, 60^@,300^@, 180^@}`
Within the unit circle

Explanation:

Here is a simple
approach

we know `cos^2 A - sin^2 A = cos 2A`

`- cosA = cos(-A)`

Using these we get;

`cos^2x-sin^2x= -cosx`

`cos 2x= cos (- x)`

`=> 2x = -x => 3x = 0 ,x = 0`

Right this is a definite solution

Lets go back to the equation

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

Bring everything over to one side

Let `cos x = a`

`2a^2 + a -1 = 0`

Factoring you get

`(2a -1)(a + 1) = 0`

`2a - 1 = 0`

`a = 1/2`

`=>cos x = 1/2`

Now The first value which comes to mind is

`x = 60^@`

Another value is

`x = 300^@`

Lets repeat the same for the other part of the equation

`a + 1 = 0`

`a = -1`

`=>cos x =- 1`

There is only 1 possibility within a unit circle

`x = 180`