Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Functions

The principal solutions of tan 3θ = –1 are

A

π4,7π12,11π12,π16,19π4,23π12\frac {π}{4}, \frac {7π}{12}, \frac {11π}{12}, \frac {π}{16}, \frac {19π}{4}, \frac {23π}{12}

B

π4,7π12,11π12,5π4,19π12,23π12\frac {π}{4}, \frac {7π}{12}, \frac {11π}{12}, \frac {5π}{4}, \frac {19π}{12}, \frac {23π}{12}

C

π4,π12\frac {π}{4}, \frac {π}{12}

D

π4,π12,13π12,7π4,19π4,23π12\frac {π}{4}, \frac {π}{12}, \frac {13π}{12}, \frac {7π}{4}, \frac {19π}{4}, \frac {23π}{12}

Answer

π4,7π12,11π12,5π4,19π12,23π12\frac {π}{4}, \frac {7π}{12}, \frac {11π}{12}, \frac {5π}{4}, \frac {19π}{12}, \frac {23π}{12}

Explanation

Solution

Tan 3θ = -1
Tan 3θ = - Tan π4\frac {π}{4} = Tan (π - π4\frac {π}{4}) = Tan (2π - π4\frac {π}{4}) = Tan (3π - π4\frac {π}{4}) = Tan (4π - π4\frac {π}{4}) = Tan (5π - π4\frac {π}{4}) = Tan (6π - π4\frac {π}{4})
Tan 3θ = tan 3π4\frac {3π}{4} = tan 7π4\frac {7π}{4} = tan 11π4\frac {11π}{4} = tan 15π4\frac {15π}{4} = tan 19π4\frac {19π}{4} = tan 23π4\frac {23π}{4}
3θ = 3π4\frac {3π}{4} = 7π4\frac {7π}{4} = 11π4\frac {11π}{4} = 15π4\frac {15π}{4} = 19π4\frac {19π}{4} = 23π4\frac {23π}{4}
θ = π4\frac {π}{4} = 7π12\frac {7π}{12} = 11π12\frac {11π}{12} =5π4\frac {5π}{4}= 19π12\frac {19π}{12} = 23π12\frac {23π}{12}
So principal solutions are {π4\frac {π}{4}, 7π12\frac {7π}{12}, 11π12\frac {11π}{12}, 5π4\frac {5π}{4}, 19π12\frac {19π}{12}, 23π12\frac {23π}{12}}
Therefore the correct option is (B).