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 9 months 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?)
 9 months 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
 9 months ago
Best ResponseYou've already chosen the best response.5\[\cos2\theta=\frac{ \sqrt{3} }{ 2 }\]

gorv
 9 months 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
 9 months 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
 9 months 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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