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Question
The number of real roots of the equationtan(x+π6)=2 tanx, for x∈(0, 3π) are
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Solution
The correct option is A \N
We are given thattan(x+π6)=2 tanx⇒tanx+tanπ61−tanx tanπ6=2 tanxLet t=tanx, we get⇒t+1√31−t√3 =2t⇒t+1√3=2t(1−t√3)⇒√3t+1=2√3t(1−t√3)⇒√3t+1=2√3t−2t2⇒2t2−√3t+1=0For the above quadratic equationDiscriminant D=3−8<0,Hence, the equation has no real solution.
We are given thattan(x+π6)=2 tanx⇒tanx+tanπ61−tanx tanπ6=2 tanxLet t=tanx, we get⇒t+1√31−t√3 =2t⇒t+1√3=2t(1−t√3)⇒√3t+1=2√3t(1−t√3)⇒√3t+1=2√3t−2t2⇒2t2−√3t+1=0For the above quadratic equationDiscriminant D=3−8<0,Hence, the equation has no real solution.
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