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Gemiola [76]
3 years ago
9

1. Solve the given equation on the domain [0,2pi]

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer:

1.x = pi/3 or 4pi/3

2. x = -pi/6 +2pi*n  or 5pi/6 + 2pi *n  where n is an integer

Step-by-step explanation:

1. sqrt(3) tan x = 3

Divide each side by sqrt(3)

sqrt(3)/sqrt(3) tan x = 3/sqrt(3)

tan x = sqrt(3) * sqrt(3)/sqrt(3)

tan x = sqrt(3)

Take the inverse of each side

arctan (tanx) = arctan (sqrt(3))

x = arctan (sqrt(3))

x =pi/3  or - 2pi/3

Since the domain is between 0 and 2pi, add 2pi to - 2pi/3  since the trig functions  are circular

x = pi/3  or -2pi/3 + 6pi/3

x = pi/3 or 4pi/3


2.  3 tan x = -sqrt(3)

Divide each side by 3

3/3 tan x = -sqrt(3)/3

tan x = -sqrt(3)/3

Take the arctan of each side

arctan (tan x) = arctan ( -sqrt(3)/3)

x = arctan ( -sqrt(3)/3)

x =-pi/6 or 5pi/6

We want all values for x so add 2pi*n to each value where n is an integer

x = -pi/6 +2pi*n  or 5pi/6 + 2pi *n  where n is an integer

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Answer:

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