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Homework Statement
Solve for 0 ≤ θ ≤ 2π (pi). Give the exact values.
tan^2θ=2tanθsinθ
I'm not sure how to go about this one =/
the other one would be 5π/3The [tex]cos(\theta) = 1/2[/tex] is correct, but I don't know how you arrived at [tex]cos(\theta) = 0[/tex]. If [tex]cos(\theta) = 0[/tex], then [tex]tan(\theta)[/tex] is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for [tex]cos(\theta) = 1/2[/tex] on [tex][0,2\pi][/tex], one of them is [tex]\pi/3[/tex] as you mentioned. What do you think the other one is? Drawing a graph out might help.
Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.I'm not sure how to go about this one =/
[tex]\cos{\theta}[/tex] doesn't work because tangent is undefines when cosθ=0. You also divided by sinθ, which assumes that [tex]\sin{\theta}\not= 0[/tex]. This caused you to lose that solution which works.the other one would be 5π/3
here's what I did
tan^2θ=2tanθsinθ
sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
1/cos^2 θ=2/cosθ
2cos^2 θ-cosθ=0
cosθ(2cosθ -1)=0
cosθ=0 and 2cosθ -1=0