solve cotθ = square root of 3 for θ
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\[\cot \theta=\sqrt{3}\] \[\frac{1}{\tan \theta}=\sqrt{3}\] \[\tan \theta=\frac{1}{\sqrt{3}}\] now let's think when we go around the unit circle will \[\frac{\sin \theta}{\cos \theta}=\frac{1}{\sqrt3}\] how about what happens at pi/6 \[\sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \cos \frac{\pi}{6}=\frac{1}{2}\] pi/6 will not wotk since tanpi/6 will not give us 1/sqrt{3} so the answer is pi/3
but this repeats it self as we go around the circle some more
cosx/sinx =root(3), so cosx = sinx * root(3), lets square both sides cos²x = 3sin²x, then sin²x = 1 - cos²x, so cos²x = 3(1-cos²x), so cos²x = 3-3cos²x, so 4cos²x = 3, then cosx = root(3)/2, special triangles and that is x = pi/3.
Maniniya is right, but that answer is derived from intuition where as mine is a systematic process, hope that helps :)
thank you very much.
Np, i wouldn't mind getting a life saver :P
this is my first time posting a question. how do I give those?
or the "good answer" button?
we also have x=7pi/6 so all the answers are x=pi/3+2npi x=7pi/6+2npi for n=0,1,2,3,4,...
I think u click on the user name and then click life saver
you can become a fan of his by clicking his name or you can click good answer to give him a medal or you can do both
where did you get "pi/6"?
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