Asked by Mark
28. Find all solutions if 0° ≤θ <360°. 4sin2
(3θ)−3= 0
(3θ)−3= 0
Answers
Answered by
Steve
I assume you mean
4sin^2(3θ) - 3 = 0
sin^2(3θ) = 3/4
sin(3θ) = √3/2
3θ = 60° or 120°
θ = 20° or 40°
Since sin(3θ) has period 360/3 = 120°, add multiples of 120° to those values. You will find six solutions in the desired domain.
4sin^2(3θ) - 3 = 0
sin^2(3θ) = 3/4
sin(3θ) = √3/2
3θ = 60° or 120°
θ = 20° or 40°
Since sin(3θ) has period 360/3 = 120°, add multiples of 120° to those values. You will find six solutions in the desired domain.
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