Asked by big145
Explain how many solutions the following trigonometric equation should have:
2cos^2x - 1 = 0, when 0 ≤ x ≤ 2𝛑
2cos^2x - 1 = 0, when 0 ≤ x ≤ 2𝛑
Answers
Answered by
oobleck
2cos^2x - 1 = cos(2x)
so 2x = (𝛑/2 or 3𝛑/2) + k*2𝛑
x = (𝛑/4 or 3𝛑/4) + k𝛑
so how many values of k produce a result in [0,2𝛑]?
so 2x = (𝛑/2 or 3𝛑/2) + k*2𝛑
x = (𝛑/4 or 3𝛑/4) + k𝛑
so how many values of k produce a result in [0,2𝛑]?
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