The values of θ satisfying sin 7θ = sin 4θ - sin θ and 0

Question

The values of θ satisfying sin 7θ = sin 4θ - sin θ and 0 < θ <(π/2) are (A) (π/9),(π/4) (B) (π/3),(π/9) (C) (π/6),(π/9) (D) (π/3),(π/4)

Tardigrade Admin · Accepted Answer

sin 7 θ= sin 4 θ- sin θ ⇒ sin 7 θ+ sin θ + sin θ= sin 4 θ ⇒ 2 sin 4 θ cos 3 θ= sin 4 θ ⇒ sin 4θ(2 cos 3θ-1)=0 ⇒ sin 4 θ=0 or cos 3θ=(1/2) 4θ=π,2π,3π.....or 3θ=(π/3),(5π/3)..... ⇒ θ=(π/4), and θ (π/9) ∈ (0,(π/2)) Thus θ=(π/4),(π/9)

Q. The values of $θ$ satisfying $sin 7 θ = sin 4 θ - sin θ$ and $0 < θ < \frac{π}{2}$ are

$\frac{π}{9}, \frac{π}{4}$

$\frac{π}{3}, \frac{π}{9}$

$\frac{π}{6}, \frac{π}{9}$

$\frac{π}{3}, \frac{π}{4}$

Q. The values of θ satisfying sin7θ=sin4θ−sinθ and 0<θ<2π​ are

9π​,4π​

3π​,9π​

6π​,9π​

3π​,4π​

sin7θ=sin4θ−sinθ ⇒sin7θ+sinθ+sinθ=sin4θ ⇒2sin4θcos3θ=sin4θ ⇒sin4θ(2cos3θ−1)=0 ⇒sin4θ=0 or cos3θ=21​ 4θ=π,2π,3π.....or3θ=3π​,35π​..... ⇒θ=4π​, and θ9π​∈(0,2π​) Thus θ=4π​,9π​

Q. The values of $θ$ satisfying $sin 7 θ = sin 4 θ - sin θ$ and $0 < θ < \frac{π}{2}$ are

$\frac{π}{9}, \frac{π}{4}$

$\frac{π}{3}, \frac{π}{9}$

$\frac{π}{6}, \frac{π}{9}$

$\frac{π}{3}, \frac{π}{4}$