Solve the equation for the interval [0°, 360°):

Question

4cos^2(theta)sin(theta)+2cos^2(theta)=0

Answer 1

is the

answer.

Comments

cos(θ) = 0 or sin(θ) = - 1/2.

θ = cos^{- 1}(0) or θ = sin^{- 1}(- 1/2).

• cos(θ) = 0.

cos(θ) = cos(π/2)

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

θ = 2nπ ± π/2

θ = 2nπ + π/2 or θ = 2nπ - π/2

The solutions outside the interval [ 0, 2π ) are

If n = 0 then θ = 2(0)π - π/2 = -π/2 < 0

If n = 1 then θ = 2(1)π + π/2 = 5π/2 > 0.

The solutions in the interval [ 0, 2π ) are

θ = 2(0)π + π/2 = π/2, and

θ = 2(1)π - π/2 = 3π/2

The solutions are π/2 and 3π/2.

• The function sin(θ) has a period of 2π, first find all solutions in the interval (0, 2π).

The function sin(θ) is negative in third and fourth quadrant.

• sin(θ) = - 1/2.

In third Quadrant, π ≤ θ ≤ 3π/2.

- 1/2 = - sin (π/6) = sin (π + π/6) = sin (7π/6).

⇒ θ = 7π/6.

In fourth Quadrant, 3π/2 ≤ θ ≤ 2π.

- 1/2 = - sin (π/6) = sin (2π - π/6) = sin (11π/6).

⇒ θ = 11π/6.

The solutions are 7π/6 and 11π/6.

The solutions in the interval [ 0, 2π ) are π/2, 3π/2, 7π/6 and 11π/6.