2cos(2x)-1=-3sin^2x; -2pi<x<2pi

Question

PLEASE ANSWER THE QUESTON!

Answer 1

The trigonometric equation is 2 cos(2x) - 1 = - 3 sin² x and solve for x.

2 cos(2x) - 1 + 3 sin² x = 0.

Apply double angle formula : cos 2u = 1 - 2sin² u.

2(1 - 2sin² x) - 1 + 3sin² x = 0

2 - 4sin² x - 1 + 3sin² x = 0

1 - sin² x = 0

cos² x = 0

cos x = 0.

The function cos x has a period of 2π, first find all solutions in the interval [0, 2π).

cos x = cos (π/2) and cos x = cos (3π/2).

x = π/2 and x = 3π/2.

Finally, add multiples of 2π to each of these solutions to get the general form

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

If n = - 1 then

x = π/2 + 2nπ -----> x = π/2 - 2π -----> x = - 3π/2 and

x = 3π/2 + 2nπ -----> x = 3π/2 - 2π -----> x = - π/2.

If n = 0 then

x = π/2 + 2nπ -----> x = π/2 - 0 -----> x = π/2 and

x = 3π/2 + 2nπ -----> x = 3π/2 - 0 -----> x = 3π/2.

If n = 1 then

x = π/2 + 2nπ -----> x = π/2 + 2π -----> x = 5π/2 and

x = 3π/2 + 2nπ -----> x = 3π/2 + 2π -----> x = 7π/2.

In the interval [-2π, 2π], the solutions are - 3π/2, - π/2, π/2 and 3π/2.