Question

The principal solutions of tan 3θ = –1 are

• A. \(\frac {π}{4}, \frac {7π}{12}, \frac {11π}{12}, \frac {π}{16}, \frac {19π}{4}, \frac {23π}{12}\)
• B. \(\frac {π}{4}, \frac {7π}{12}, \frac {11π}{12}, \frac {5π}{4}, \frac {19π}{12}, \frac {23π}{12}\)
• C. \(\frac {π}{4}, \frac {π}{12}\)
• D. \(\frac {π}{4}, \frac {π}{12}, \frac {13π}{12}, \frac {7π}{4}, \frac {19π}{4}, \frac {23π}{12}\)

Answer 1

The Correct Option is B

Tan 3θ = -1
Tan 3θ = - Tan \(\frac {π}{4}\) = Tan (π - \(\frac {π}{4}\)) = Tan (2π - \(\frac {π}{4}\)) = Tan (3π - \(\frac {π}{4}\)) = Tan (4π - \(\frac {π}{4}\)) = Tan (5π - \(\frac {π}{4}\)) = Tan (6π - \(\frac {π}{4}\))
Tan 3θ = tan \(\frac {3π}{4}\) = tan \(\frac {7π}{4}\) = tan \(\frac {11π}{4}\) = tan \(\frac {15π}{4}\) = tan \(\frac {19π}{4}\) = tan \(\frac {23π}{4}\)
3θ = \(\frac {3π}{4}\) = \(\frac {7π}{4}\) = \(\frac {11π}{4}\) = \(\frac {15π}{4}\) = \(\frac {19π}{4}\) = \(\frac {23π}{4}\)
θ = \(\frac {π}{4}\) = \(\frac {7π}{12}\) = \(\frac {11π}{12}\) =\(\frac {5π}{4}\)= \(\frac {19π}{12}\) = \(\frac {23π}{12}\)
So principal solutions are {\(\frac {π}{4}\), \(\frac {7π}{12}\), \(\frac {11π}{12}\), \(\frac {5π}{4}\), \(\frac {19π}{12}\), \(\frac {23π}{12}\)}
Therefore the correct option is (B).