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vladimir2022 [97]
3 years ago
10

2 cot^2t sin t - cot^2t = 0

Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

2 cot²(<em>t </em>) sin(<em>t </em>) - cot²(<em>t </em>) = 0

cot²(<em>t </em>) (2 sin(<em>t </em>) - 1) = 0

cot²(<em>t </em>) = 0   <u>or</u>   2 sin(<em>t</em> ) - 1 = 0

cot(<em>t</em> ) = 0   <u>or</u>   sin(<em>t</em> ) = 1/2

cos(<em>t</em> ) / sin(<em>t</em> ) = 0   <u>or</u>   sin(<em>t</em> ) = 1/2

cos(<em>t</em> ) = 0   <u>or</u>   sin(<em>t</em> ) = 1/2

[<em>t</em> = cos⁻¹(0) + 2<em>nπ</em>   <u>or</u>   <em>t</em> = cos⁻¹(0) - <em>π</em> + 2<em>nπ</em>]

  <u>or</u>   [<em>t</em> = sin⁻¹(1/2) + 2<em>nπ</em>   <u>or</u>   <em>t</em> = <em>π</em> - sin⁻¹(1/2) + 2<em>nπ</em>]

(where <em>n</em> is any integer)

<em>t</em> = <em>π</em>/2 + 2<em>nπ</em>   <u>or</u>   <em>t</em> = -<em>π</em>/2 + 2<em>nπ</em>   <u>or</u>   <em>t</em> = <em>π</em>/6 + 2<em>nπ</em>   <u>or</u>   <em>t</em> = 5<em>π</em>/6 + 2<em>nπ</em>

Note that the first two families of solutions overlap and can be condensed, so that

<em>t</em> = <em>π</em>/2 + <em>nπ</em>   <u>or</u>   <em>t</em> = <em>π</em>/6 + 2<em>nπ</em>   <u>or</u>   <em>t</em> = 5<em>π</em>/6 + 2<em>nπ</em>

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