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

2 Answers
Konstantinos Michailidis
Mar 3, 2016

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)#

Sidharth
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#

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