4 tan(3θ) - 4 = 0

4 tan(3θ) - 4 = 0tan(3θ) = 4/4 =1tan(3θ) =13θ = tan^-1 (1) 3θ = π/4, 5π/4 , 9π/4 , 13π/4 , 17π/4 , 21π/4θ in the interval of [0,2π) is π/12,5π/12,3π/4 , 13π/12,17π/12 and 21π/12Let me know if you have any doubts.

is it ok or you looking for something else??

It is good but seems to be lacking some steps

Its hard to give all the steps here....I hope you understand.but its mainly, tan^-1 (1) = π/4+ kπ , where K= 0,1,2,3,4so, you get the answers by putting K=0,1,2,3.....If it was helpful then please pay me...

Thank you but I just selected another answer

sorry

oh okay nevermind :)

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