## 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?) 9 months ago 9 months ago

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1. gorv

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

2. gorv

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

3. gorv

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

4. Satsuki

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