-cosx=-sin^2 x-1 find all solution in interval [0,2pi)

Question

CAN SOMEONE PLEASE HELP ME FOR THIS QUESTION

 

Answer 1

Saim,

 

Your problem is to solve

 

− cos x = − sin² x − 1 for all solutions is the interval [0, 2π).

 

Since you have two different functions (cos and sin), the quickest way to do this is to rewrite the equation in terms of only one of them. This is best done using the identity

 

sin² x = 1 − cos² x.

 

After you plug this in, you have

 

− cos x = − (1 − cos² x) − 1.

 

Now, simplify the algebra.

 

− cos x = −1 + cos² x − 1

 

− cos x = cos² x − 2

 

0 = cos² x + cos x − 2.

 

You can now try factoring the right sides:

 

0 = (cos x + 2)(cos x − 1)

 

Then, set each factor equal to zero and solve for cos x:

 

cos x = −2  or  cos x = 1.

 

The first of these has no solution (why?) and the second has only one in [0,2π). It is x = 0.

 

You can check this by plugging into the original equation. the left side is −1 and the right side is −0² − 1, which eqauls the same number.

 

The solution is x = 0.