In the interval [0,2π), find the value of B: 3tan^2(3x) = 3?

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answers:
la console say: 3.tan²(3x) = 3

tan²(3x) = 1

tan(3x) = ± 1


First case:

tan(3x) = 1 ← the corresponding angle is: π/4

3x = (π/4) + kπ → where k is an integer

x = (π/12) + k.(π/3)


When: k = 0 → x = π/12

When: k = 1 → x = (π/12) + (π/3) → x = 5π/12

When: k = 2 → x = (π/12) + (2π/3) → x = 3π/4

When: k = 3 → x = (π/12) + (3π/3) → x = 13π/12

When: k = 4 → x = (π/12) + (4π/3) → x = 17π/12

When: k = 5 → x = (π/12) + (5π/3) → x = 7π/4

When: k = 6 → x = (π/12) + (6π/3) → x = 25π/12 ← over 2π


Second case:

tan(3x) = - 1 ← the corresponding angle is: 3π/4

3x = (3π/4) + kπ → where k is an integer

x = (π/4) + k.(π/3)


When: k = 0 → x = π/4

When: k = 1 → x = (π/4) + (π/3) → x = 7π/12

When: k = 2 → x = (π/4) + (2π/3) → x = 11π/12

When: k = 3 → x = (π/4) + (3π/3) → x = 5π/4

When: k = 4 → x = (π/4) + (4π/3) → x = 19π/12

When: k = 5 → x = (π/4) + (5π/3) → x = 23π/12

When: k = 6 → x = (π/4) + (6π/3) → x = 9π/4 ← over 2π


→ Solution = { π/12 ; 5π/12 ; 3π/4 ; 13π/12 ; 17π/12 ; 7π/4 ; π/4 ; 7π/12 ; 11π/12 ; 5π/4 ; 19π/12 ; 23π/12 }
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MyRank say: 3tan²3x = 3

tan²3x = 3/3

tan²3x = 1

tan3x = ±1

tan3x = 1

3x = tan⁻¹(1)

3x = π/4

x = π/12

tan3x = -1

3x = tan⁻¹(-1)

3x = π+π/4.
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Pope say: You would first have to define B. That character appears nowhere in your given condition.
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Captain Matticus, LandPiratesInc say: 3 * tan(3x)^2 = 3
tan(3x)^2 = 1
tan(3x) = -1 , 1
3x = pi/4 + pi * k , 3pi/4 + pi * k
3x = pi/4 + (pi/2) * k
3x = (pi/4) * (1 + 2k)
x = (pi/12) * (1 + 2k)

x = pi/12 , 3pi/12 , 5pi/12 , 7pi/12 , 9pi/12 , 11pi/12 , 13pi/12 , 15pi/12 , 17pi/12 , 19pi/12 , 21pi/12 , 23pi/12
x = pi/12 , pi/4 , 5pi/12 , 7pi/12 , 3pi/4 , 11pi/12 , 13pi/12 , 5pi/4 , 17pi/12 , 19pi/12 , 7pi/4 , 23pi/12
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