Question

Solve the equation in the interval [0, 2π). cos 2x = cos x - 1.

Answer 1

The trigonometric equatino is cos 2x = cos x - 1.

Rewrite the equation as cos^2 (x ) - sin^2 (x ) = cos x - [ sin^2 (x ) + cos^2 (x ) ].

cos^2 (x ) - sin^2 (x ) = cos x - sin^2 (x ) - cos^2 (x )

2 cos^2 (x ) - cos x = 0

cos x (2 cos x - 1) = 0

Set each factor equal to zero to find the solutions in the interval [0, 2π).

cos x = 0                 and                2cos x - 1 = 0

       x = π / 2, 3π / 2.                          2cos x = 1

                                                             cos x = 1 / 2

                                                                    x = π / 3, 5π / 3.

Because cos x has a period of 2π, the general form of the solution is obtained by adding multiples of 2π to get the general solution, i.e,

x = π / 2 + 2nπ, x = 3π / 2 + 2nπ, x = π / 3 + 2nπ, and x = 5π / 3 + 2nπ, where n is an integer.