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Find all solutions in the interval [0,2π)

Calculus
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4 tan(3θ) - 4 = 0

Mar 17th, 2015

4 tan(3θ) - 4 = 0
tan(3θ) = 4/4 =1

tan(3θ) =1
3θ = tan^-1 (1)
3θ = π/4, 5π/4 , 9π/4 , 13π/4 , 17π/4 , 21π/4

θ in the interval of [0,2π) is π/12,5π/12,3π/4 , 13π/12,17π/12 and 21π/12


Let me know if you have any doubts.

Mar 17th, 2015

is it ok or you looking for something else??

Mar 17th, 2015

It is good but seems to be lacking some steps


Mar 17th, 2015

Its hard to give all the steps here....I hope you understand.

but its mainly,  tan^-1 (1) = π/4+ kπ , where K= 0,1,2,3,4

so, you get the answers by putting K=0,1,2,3.....


If it was helpful then please pay me...

Mar 17th, 2015

Thank you but I just selected another answer

sorry

Mar 17th, 2015

oh okay nevermind :)

Mar 17th, 2015

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