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?)

Question

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?)

gorv · Answer

$$\cos2\theta=-\frac{ \sqrt{3} }{ 2 }$$

gorv · Answer

$$2\theta= \pi+ \cos^-1\frac{ \sqrt{3} }{ 2 }  or =\pi-\cos^-1\frac{ \sqrt{3} }{ 2 }$$

gorv · Answer

two solution bcozzz cos gives negative value only in 2nd quadrent
that is pi-angle
or in third quadarent that is pi + angle
2 theta=pi +pi/3 .....or 2 theta= pi-pi/3
theta = (4pi/3)/2....or theta=(2pi/3)/2
theta=2pi/3......or theta=pi/3

satsuki · Answer

If I follow correctly, then that means the solution set will only be {5π/12 , 17π/12}, correct?