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Naddik [55]
3 years ago
5

Find the sixth term of an arithmetic sequence with t1 = 2 and tn = tn-1 + 4.

Mathematics
1 answer:
ZanzabumX [31]3 years ago
8 0
 <span>The sixth term of an arithmetic sequence with t1 = 2 and tn = tn-1 + 4 is 22. </span><span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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What are the values that satisfy the trigonometric equation for: 0 ≤ q ≤ 2Π ?
bagirrra123 [75]
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.
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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