Maths - Please Help?

Question

Hello, 

There are some questions that I am not sure how to tackle: 

1) 3cos2θ - cosθ + 2 = 0 

2) tan2θ + tanθ = 0 

3) 2cos2x = 5 - 13sinx 

The answers are apparently: 

1) (pi)/3, [5(pi)]/3, 1.91, 4.37 

2) 0, (pi)/3, [2(pi)]/3, (pi), [4(pi)]/3, [5(pi)]/3, 2(pi) 

3) 0, [4(pi)]/3, 2(pi) 

Answer 1

1).

The trigonometric equation is 3cos 2θ - cos θ + 2 = 0.

Double angle formula : cos(2θ) = 2cos² θ - 1.

3cos 2θ - cos θ + 2 = 0

3(2cos² θ - 1) - cos θ + 2 = 0

6cos² θ - 3 - cos θ + 2 = 0

6cos² θ - cos θ - 1 = 0

6cos² θ - 3cos θ + 2cos θ - 1 = 0

3cos θ(2cos θ - 1) + 1(2cos θ - 1) = 0

(2cos θ - 1)(3cos θ + 1) = 0

2cos θ - 1 = 0  and  3cos θ + 1 = 0

cos θ = 1/2  and  cos θ = - 1/3

• cos θ = 1/2.

cos (θ) = cos(π/3)

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

⇒θ = 2nπ ± (π/3)

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

If n = 1, θ = 2(1)π + (π/3) and θ = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

⇒ θ = π/3 and θ = 5π/3 in the interval [0, 2π).

• cos θ = - 1/3.

cos (θ) = cos(cos^{- 1}(- 1/3))

cos (θ) = cos(1.91)

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

⇒θ = 2nπ ± 1.91

If n = 0, θ = 2(0)π + 1.91 and θ = 2(0)π - 1.91 = 1.91 and - 1.91,

If n = 1, θ = 2(1)π + 1.91 and θ = 2(1)π - 1.91 = 2π + 1.91 and 2π - 1.91 = 8.91 and 4.37.

⇒ θ = 1.91 and θ = 4.37 in the interval [0, 2π).

Therefore, the solutions are θ = π/3, θ = 5π/3, θ = 1.91 and θ = 4.37 in the interval [0, 2π).

Answer 2

3).

The trigonometric equation is 2cos 2x = 5 - 13sin x.

Double angle formula : cos(2x) = 1 - 2sin² x.

2(1 - 2sin² x) = 5 - 13sin x

2 - 4sin² x - 5 + 13sin x = 0

4sin² x - 13sin x + 3 = 0

4sin² x - 12sin x - sin x + 3 = 0

4sin x(sin x - 3) - 1(sin x - 3) = 0

(sin x - 3)(4sin x - 1) = 0

sin x = 3  and  sin x = 1/4

• sin x = 3.

The range of the sine function is [- 1, 1].

In this case : sin x = 3, x does not exists.

• sin x = 1/4.

sin x = sin(sin^{- 1}(1/4))

sin x = sin(0.0804π).

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

⇒ x = nπ + (- 1)ⁿ(0.0804π).

The solution is x = nπ + (- 1)ⁿ(0.0804π), where n is an integer.

Answer 3

2).

The trigonometric equation is tan 2θ + tan θ = 0.

Double angle formula : tan(2θ) = 2tan θ/(1 - tan² θ).

[2tan θ/(1 - tan² θ)] + tan θ = 0

2tan θ + tan θ(1 - tan² θ) = 0

tan θ(2 + 1 - tan² θ) = 0

tan θ(3 - tan² θ) = 0

tan θ = 0  and  3 - tan² θ = 0

tan θ = 0  and  tan² θ = 3

tan θ = 0  and  tan θ = ± √3.

• tan θ = 0.

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

⇒ θ = nπ.

If, n = 0, then θ = 0*π = 0,

If, n = 1, then θ = 1*π = π,

If, n = 2, then θ = 2*π = 2π,....etc

• tan θ = - √3

tan θ = tan(- π/3).

⇒ θ = nπ - π/3.

If, n = 0, then θ = 0*π - π/3 = - π/3,

If, n = 1, then θ = 1*π - π/3 = (3π - π)/3 = 2π/3,

If, n = 2, then θ = 2*π - π/3 = (6π - π)/3 = 5π/3, .....etc.

• tan θ = √3

tan θ = tan(π/3).

⇒ θ = nπ + π/3.

If, n = 0, then θ = 0*π + π/3 = π/3,

If, n = 1, then θ = 1*π + π/3 = (3π + π)/3 = 4π/3,

If, n = 2, then θ = 2*π + π/3 = (6π + π)/3 = 7π/3, .....etc.

Therefore, θ = 0, π/3, 2π/3, π, 4π/3, 5π/3, and 2π.