Asked by Mariam
So I'm given a general equation and am expected to get the location on the unit circle, the period, and the general solution to it. The equation is (3cot^2)2x-1 = 0. I've gotten it down to (cot^2)2x = 1/3 but I have no clue what to do next. Please help!
Answers
Answered by
Reiny
ok so far
(cot^2)2x = 1/3
tan^2 (2x) = 3
tan (2x) = ±√3
I recall that tan 60° = √3
so 2x = 60
2x = 120° in quad II
2x = 240° in quad III
2x = 300° in quad IV
so x = 30°, 60°, 120°, or 150° for 0 ≤ x ≤ 360°
in radians: x = π/6, π/3, 2π/3, 5π/3
take it from there
(cot^2)2x = 1/3
tan^2 (2x) = 3
tan (2x) = ±√3
I recall that tan 60° = √3
so 2x = 60
2x = 120° in quad II
2x = 240° in quad III
2x = 300° in quad IV
so x = 30°, 60°, 120°, or 150° for 0 ≤ x ≤ 360°
in radians: x = π/6, π/3, 2π/3, 5π/3
take it from there
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