MATH SOLVE

4 months ago

Q:
# What are the values that satisfy the trigonometric equation for: 0 ≤ q ≤ 2Π ?Sinθ + tan(-θ) = 0 (\pi )/(2), (3\pi )/(2)0, \pi (\pi )/(4),(5\pi )/(4)0, (\pi )/(2), \pi , (3\pi )/(2)

Accepted Solution

A:

Answer: 0, π, 2π

Nevertheless, that is not an option. I see two possibilities: 1) the options are misswirtten, 2) the domain is not well defined.

If the domain were 0 ≤ θ < 2π, then 2π were excluded of the domain ant the answer would be 0, π.

Explanation:

1) The first solution, θ = 0 is trivial:

sin (0) - tan (0) = 0

0 - 0 = 0

2) For other solutions, work the expression:

sin(θ) + tan (-θ) = 0 ← given

sin (θ) - tan(θ) = 0 ← tan (-θ) = tan(θ)

sin(θ) - sin (θ) / cos(θ) = 0 ← tan(θ) = sin(θ) / cos(θ)

sin (θ) [1 - 1/cos(θ)] = 0 ← common factor sin(θ)

⇒ Any of the two factors can be 0

⇒ sin (θ) = 0 or (1 - 1 / cos(θ) = 0,

sin(θ) = 0 ⇒ θ = 0, π, 2π

1 - 1/cos(θ) = 0 ⇒ 1/cos(θ) = 1 ⇒ cos(θ) = 1 ⇒ θ = 0, 2π

⇒ Solutions are 0, π, and 2π

In fact if you test with any of those values the equation is checked. The only way to exclude one of those solutions is changing the domain.

Nevertheless, that is not an option. I see two possibilities: 1) the options are misswirtten, 2) the domain is not well defined.

If the domain were 0 ≤ θ < 2π, then 2π were excluded of the domain ant the answer would be 0, π.

Explanation:

1) The first solution, θ = 0 is trivial:

sin (0) - tan (0) = 0

0 - 0 = 0

2) For other solutions, work the expression:

sin(θ) + tan (-θ) = 0 ← given

sin (θ) - tan(θ) = 0 ← tan (-θ) = tan(θ)

sin(θ) - sin (θ) / cos(θ) = 0 ← tan(θ) = sin(θ) / cos(θ)

sin (θ) [1 - 1/cos(θ)] = 0 ← common factor sin(θ)

⇒ Any of the two factors can be 0

⇒ sin (θ) = 0 or (1 - 1 / cos(θ) = 0,

sin(θ) = 0 ⇒ θ = 0, π, 2π

1 - 1/cos(θ) = 0 ⇒ 1/cos(θ) = 1 ⇒ cos(θ) = 1 ⇒ θ = 0, 2π

⇒ Solutions are 0, π, and 2π

In fact if you test with any of those values the equation is checked. The only way to exclude one of those solutions is changing the domain.