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Solve on the interval [0, 2pi] 3 sec x - 1 = 2​

User Amazingred
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Explanation:

3 sec(x) - 1 = 2

Adding 1 to both sides:

3 sec(x) = 3

Dividing by 3:

sec(x) = 1

Taking the inverse secant of both sides:

x = cos^-1(1)

Since cosine has a period of 2π, the solutions to this equation occur at:

x = 0 + 2πn and x = 2π - 2πn, where n is an integer.

Substituting in n = 0, we get:

x = cos^-1(1) = 0

Substituting in n = 1, we get:

x = 2π - cos^-1(1) = 2π

Therefore, the solutions on the interval [0, 2π] are:

x = 0 and x = 2π.

User Tinkertime
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