Solving Trig Equations?

Question

I have four problems that I just can't seem to figure out. The answers need to be rounded to 3 decimal places when needed.

1) 3 sec^2 θ − 4 = 0

2) (tan^2 θ − 16)(2 cos θ + 1) = 0

3) 7 sin^2 θ − 29 sin θ + 4 = 0

4)7 tan θ sin θ − 6 tan θ = 0

Any help is greatly appreciated!

Answer 1

(1).

The trigonometric equation is 3 sec²(θ) - 4 = 0 and solve for θ.

sec²(θ) = 4/3

sec(θ) = ± 2/√3

sec(θ) = ± sec(π/6)

sec(θ) = sec(± π/6).

General solution : If sec(θ) = sec(∝), then θ = nπ + (-1)ⁿ∝, where n is an integer.

If ∝ = ± π/6, then θ = nπ ± (-1)ⁿ(π/6), where n is an integer.

Answer 2

(2).

The trigonometric equation is [tan²(θ) - 16][2 cos(θ) + 1] = 0 and solve for θ.

Apply Zero product property : If ab = 0, then either a = 0, b = 0 or both a and b are zeros.

tan²(θ) - 16 = 0 and 2 cos(θ) + 1 = 0

Solve equation 1 : tan²(θ) - 16 = 0 for θ.

tan(θ) = ± 4

tan(θ) = ± tan[tan^-1(4)]

tan(θ) = ± tan(75.964^o)

tan(θ) = tan(± 75.964^o)

General solution : If tan(θ) = tan(∝), then θ = 180^on + ∝, where n is an integer.

If ∝ = ± 75.964^o, then θ = 180^on ± 75.964^o, where n is an integer.

 

Solve equation 2 : 2 cos(θ) + 1 = 0 for θ.

cos(θ) = - 1/2

cos(θ) = - cos(60^o)

cos(θ) = cos(- 60^o)

General solution : If cos(θ) = cos(∝), then θ = 360^on ± ∝, where n is an integer.

If ∝ = - 60^o, then θ = 360^on ± 60^o, where n is an integer.

The general solution of the original equation is θ = 180^on ± 75.964^o and θ = 360^on ± 60^o, where n is an integer.

Answer 3

(3).

The trigonometric equation is 7 sin²(θ) - 29 sin(θ) + 4 = 0 and solve for θ.

7 sin²(θ) - 28 sin(θ) - sin(θ) + 4 = 0

7 sin(θ)[sin(θ) - 4] - 1[sin(θ) - 4] = 0

[7 sin(θ) - 1][sin(θ) - 4] = 0

Apply Zero product property : If ab = 0, then either a = 0, b = 0 or both a and b are zeros.

7 sin(θ) - 1 = 0 and sin(θ) - 4 = 0

Solve equation 1 : 7 sin(θ) - 1 = 0 for θ.

sin(θ) = 1/7 = 0.143

sin(θ) = sin[sin^-1(0.143)]

sin(θ) = sin[8.213^o]

General solution : If sin(θ) = sin(∝), then θ = 180^on + (-1)ⁿ∝, where n is an integer.

If ∝ = 8.213^o, then θ = 180^on + (-1)ⁿ 8.213^o, where n is an integer.

 

Solve equation 2 : sin(θ) - 4 = 0 for θ.

sin(θ) = 4

The above equation has no solution, since the value of sin(θ) lies in the interval [-1, 1].

The general solution of the original equation is θ = 180^on + (-1)ⁿ 8.213^o, where n is an integer.

Answer 4

(4).

The trigonometric equation is 7 tan(θ) sin(θ) - 6 tan(θ) = 0 and solve for θ.

tan(θ) [7 sin(θ) - 6] = 0

Apply Zero product property : If ab = 0, then either a = 0, b = 0 or both a and b are zeros.

tan(θ) = 0 and 7 sin(θ) - 6 = 0

Solve equation 1 : tan(θ) = 0 for θ.

tan(θ) = tan(0^o)

General solution : If tan(θ) = tan(∝), then θ = 180^on + ∝, where n is an integer.

If ∝ = 0^o, then θ = 180^on, where n is an integer.

Solve equation 2 : 7 sin(θ) - 6 = 0 for θ.

sin(θ) = 6/7 = 0.857

sin(θ) = sin[sin^-1(0.857)]

sin(θ) = sin(58.997^o)

General solution : If sin(θ) = sin(∝), then θ = 180^on + (-1)ⁿ∝, where n is an integer.

If ∝ = 58.997^o, then θ = 180^on + (-1)ⁿ 58.997^o, where n is an integer.

 

The general solution of the original equation is θ = 180^on and θ = 180^on + (-1)ⁿ 58.997^o, where n is an integer.