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Solve the trigonometric equation on the interval [0,2π)
2cos2θ = √(3)
I did θ=5π/12 + kπ and θ=π/12 + kπ and
I got { π/12 , 5π/12, 13π/12, 17π/12} as my solution set but when I plug them back in and check on my calculator they don't all equal √3, what mistakes did I make? (Or is this right?)
 4 months ago
 4 months ago
Satsuki Group Title
Solve the trigonometric equation on the interval [0,2π) 2cos2θ = √(3) I did θ=5π/12 + kπ and θ=π/12 + kπ and I got { π/12 , 5π/12, 13π/12, 17π/12} as my solution set but when I plug them back in and check on my calculator they don't all equal √3, what mistakes did I make? (Or is this right?)
 4 months ago
 4 months ago

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gorv Group TitleBest ResponseYou've already chosen the best response.3
\[\cos2\theta=\frac{ \sqrt{3} }{ 2 }\]
 4 months ago

gorv Group TitleBest ResponseYou've already chosen the best response.3
\[2\theta= \pi+ \cos^1\frac{ \sqrt{3} }{ 2 } or =\pi\cos^1\frac{ \sqrt{3} }{ 2 }\]
 4 months ago

gorv Group TitleBest ResponseYou've already chosen the best response.3
two solution bcozzz cos gives negative value only in 2nd quadrent that is piangle or in third quadarent that is pi + angle 2 theta=pi +pi/3 .....or 2 theta= pipi/3 theta = (4pi/3)/2....or theta=(2pi/3)/2 theta=2pi/3......or theta=pi/3
 4 months ago

Satsuki Group TitleBest ResponseYou've already chosen the best response.0
If I follow correctly, then that means the solution set will only be {5π/12 , 17π/12}, correct?
 4 months ago
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