The most general solution of sin θ=-(1/2) and tan θ=(1/√3) are

Question

The most general solution of sin θ=-(1/2) and tan θ=(1/√3) are (A) 2nπ+(π/6) (B) 2nπ+(11π/6) (C) 2nπ+(7π/6) (D) 2nπ+(π/4)

Tardigrade Admin · Accepted Answer

sin θ=-(1/2) and tan θ=(1/√3) ∴ θ lies in the IIIrd quadrant. ∴ sin θ= sin (π+(π/6))= sin (7π/6) cos θ= cos (π+(π/6))= cos (7π/6) tan θ= tan (7 π/6) ∴ θ=2nπ+(7π/6)

Q. The most general solution of $sin θ = - \frac{1}{2}$ and $tan θ = \frac{1}{3}$ are

$2 nπ + \frac{π}{6}$

$2 nπ + \frac{11 π}{6}$

$2 nπ + \frac{7 π}{6}$

$2 nπ + \frac{π}{4}$

$sin θ = - \frac{1}{2}$ and $tan θ = \frac{1}{3}$
$∴ θ$ lies in the IIIrd quadrant.
$∴ sin θ = sin (π + \frac{π}{6}) = sin \frac{7 π}{6}$
$cos θ = cos (π + \frac{π}{6}) = cos \frac{7 π}{6}$
$tan θ = tan \frac{7 π}{6} ∴ θ = 2 nπ + \frac{7 π}{6}$

Q. The most general solution of sinθ=−21​ and tanθ=3​1​ are

2nπ+6π​

2nπ+611π​

2nπ+67π​

2nπ+4π​