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

2 Answers

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)

Mar 3, 2016

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