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cricket20 [7]
2 years ago
10

2〖sen〗^2 x+3 senx+1=0

Mathematics
1 answer:
KonstantinChe [14]2 years ago
7 0

2 sin²(<em>x</em>) + 3 sin(<em>x</em>) + 1 = 0

(2 sin(<em>x</em>) + 1) (sin(<em>x</em>) + 1) = 0

2 sin(<em>x</em>) + 1 = 0   OR   sin(<em>x</em>) + 1 = 0

sin(<em>x</em>) = -1/2   OR   sin(<em>x</em>) = -1

The first equation gives two solution sets,

<em>x</em> = sin⁻¹(-1/2) + 2<em>nπ</em> = -<em>π</em>/6 + 2<em>nπ</em>

<em>x</em> = <em>π</em> - sin⁻¹(-1/2) + 2<em>nπ</em> = 5<em>π</em>/6 + 2<em>nπ</em>

(where <em>n</em> is any integer), while the second equation gives

<em>x</em> = sin⁻¹(-1) + 2<em>nπ</em> = -<em>π</em>/2 + 2<em>nπ</em>

2 cot(<em>x</em>) sec(<em>x</em>) + 2 sec(<em>x</em>) + cot(<em>x</em>) + 1 = 0

2 sec(<em>x</em>) (cot(<em>x</em>) + 1) + cot(<em>x</em>) + 1 = 0

(2 sec(<em>x</em>) + 1) (cot(<em>x</em>) + 1) = 0

2 sec(<em>x</em>) + 1 = 0   OR   cot(<em>x</em>) + 1 = 0

sec(<em>x</em>) = -1/2   OR   cot(<em>x</em>) = -1

cos(<em>x</em>) = -2   OR   tan(<em>x</em>) = -1

The first equation has no (real) solutions, since -1 ≤ cos(<em>x</em>) ≤ 1 for all (real) <em>x</em>. The second equation gives

<em>x</em> = tan⁻¹(-1) + <em>nπ</em> = -<em>π</em>/4 + <em>nπ</em>

<em />

sin(<em>x</em>) cos²(<em>x</em>) = sin(<em>x</em>)

sin(<em>x</em>) cos²(<em>x</em>) - sin(<em>x</em>) = 0

sin(<em>x</em>) (cos²(<em>x</em>) - 1) = 0

sin(<em>x</em>) (-sin²(<em>x</em>)) = 0

sin³(<em>x</em>) = 0

sin(<em>x</em>) = 0

<em>x</em> = sin⁻¹(0) + 2<em>nπ</em> = 2<em>nπ</em>

<em />

2 cos²(<em>x</em>) + 2 sin(<em>x</em>) - 12 = 0

2 (1 - sin²(<em>x</em>)) + 2 sin(<em>x</em>) - 12 = 0

-2 sin²(<em>x</em>) + 2 sin(<em>x</em>) - 10 = 0

sin²(<em>x</em>) - sin(<em>x</em>) + 5 = 0

Using the quadratic formula, we get

sin(<em>x</em>) = (1 ± √(1 - 20)) / 2 = (1 ± √(-19)) / 2

but the square root contains a negative number, which means there is no real solution.

2 csc²(<em>x</em>) + cot²(<em>x</em>) - 3 = 0

2 (cot²(<em>x</em>) + 1) + cot²(<em>x</em>) - 3 = 0

3 cot²(<em>x</em>) - 1 = 0

cot²(<em>x</em>) = 1/3

tan²(<em>x</em>) = 3

tan(<em>x</em>) = ± √3

<em>x</em> = tan⁻¹(√3) + <em>nπ</em>  OR   <em>x</em> = tan⁻¹(-√3) + <em>nπ</em>

<em>x</em> = <em>π</em>/3 + <em>nπ</em>   OR   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

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Answer:

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