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Question

The general solution of the equation tan2α+23tanα = 1 is given by

A
θ=π2
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B
(n+12)π
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C
(6n+1)π12
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D
nπ12
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Solution

The correct option is A (6n+1)π12
tan2α+23tanα=1

(tanα+3)2 =4

tanα=23,23

or tanα=tan15, tan75 = tanπ12, tan(5π12)

α=nπ+π12=2nπ2+π12= Even π2+π12.................(1)

α=nπ+(5π12)=nπ6π12+π12

=(2n1)π2+π12=oddπ2+π12....................(2)

From(1) and (2),

α=kπ2+π12=(6k+1)π12.

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