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 one year ago
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?)
 one year ago
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?)

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gorv
 one year ago
Best ResponseYou've already chosen the best response.5\[\cos2\theta=\frac{ \sqrt{3} }{ 2 }\]

gorv
 one year ago
Best ResponseYou've already chosen the best response.5\[2\theta= \pi+ \cos^1\frac{ \sqrt{3} }{ 2 } or =\pi\cos^1\frac{ \sqrt{3} }{ 2 }\]

gorv
 one year ago
Best ResponseYou've already chosen the best response.5two 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

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