Question

Trig equations help?

Answer 1

2).

The trigonometric equation is cot(2θ - π/2) = 1.

Rewrite the equation as cot(-(π/2 - 2θ)) = 1.

Remember that cot( - θ) = - cot(θ).

- cot(π/2 - 2θ) = 1

tan(2θ) = - 1

tan (2θ) = tan (- π/4).

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

2θ = nπ - π/4

⇒ θ = nπ/2 - π/8

If n = 0, θ = (0)π/2 - π/8 = - π/8,

If n = 1, θ = (1)π/2 - π/8 = π/2 - π/8 = (4π - π)/8 = 3π/8,

If n = 2, θ = (2)π/2 - π/8 = π - π/8 = (8π - π)/8 = 7π/8,

If n = 3, θ = (3)π/2 - π/8 = 3π/2 - π/8 = (12π - π)/8 = 11π/8,

If n = 4, θ = (4)π/2 - π/8 = 4π/2 - π/8 = (16π - π)/8 = 15π/8.

Therefore, the solutions of the given equation are x = 3π/8, x = 7π/8, x = 11π/8, and x = 15π/8 in the interval [0, 2π).

3). The trigonometric equation is sec (3θ/2) = - √2.

Reciprocal identity : sec θ = 1/cos θ.

1/cos (3θ/2) = - √2

⇒ cos (3θ/2) = - 1/√2

cos (3θ/2) = cos(3π/4).

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

3θ/2 = 2nπ ± (3π/4)

3θ = 4nπ ± (3π/2)

⇒ θ = (4nπ/3) ± (π/2).

If n = 0, θ = (4*0*π/3) + (π/2) and θ = (4*0*π/3) - (π/2) = π/2 and - π/2,

If n = 1, θ = (4*1*π/3) + (π/2) and θ = (4*1*π/3) - (π/2) = 4π/3 + π/2 and θ = 4π/3 - π/2 = 11π/6 and 5π/6.

Therefore, the solutions of the given equation are x = π/2, x = 5π/6 and x = 11π/6 in the interval [0, 2π).