If sin 3 - - sin- how many solutions exist such that -2- - - - 2- -

We have, sin 3θ = sinθ ⇒sin 3θ sinθ = 0 ⇒ 2cos ( 3θ+θ/2 ) sin ( 3θ θ/2 ) = 0 ⇒cos 2 θ. sinθ = 0 ⇒cos 2 θ = 0 or sinθ = 0 ⇒θ = nπ±π/4 or θ = nπ, n ∈ I ∴θ = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

General Solution of cos theta = cos alpha

If sin 3 θ = ...

Question

If $sin 3 θ = sin θ,$ how many solutions exist such that $- 2 π < θ < 2 π$ ?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 11
We have, $sin 3 θ = sin θ$
$\Rightarrow sin 3 θ - sin θ = 0$
$\Rightarrow 2 cos (\frac{3 θ + θ}{2}) sin (\frac{3 θ - θ}{2}) = 0$
$\Rightarrow cos 2 θ . sin θ = 0$
$\Rightarrow cos 2 θ = 0$ or $sin θ = 0$
$\Rightarrow θ = n π \pm \frac{π}{4} o r θ = n π, n \in I$
$∴ θ = \pm \frac{π}{4}, \pm \frac{3 π}{4}, \pm \frac{5 π}{4}, \pm \frac{7 π}{4}, 0, \pm π$
$(∵ - 2 π$ $< θ < 2 π)$
Thus, total number of solutions = 11

Suggest Corrections

Similar questions

Q. If

sin 3 θ = sin θ,

how many solutions exist such that

- 2 π < θ < 2 π

Q. If

sin 3 θ = sin θ,

how many solutions exist such that

- 2 π < θ < 2 π

\sqrt{3} cos θ + sin θ = 1; - 2 π < θ < 2 π

Q. Prove the following.

(sin θ + cos θ)^{n} = cos n (\frac{1}{2} π - θ) + i sin n (\frac{1}{2} π - θ)

Q.
If

γ sin θ = 3, γ = 4 (1 + sin θ), 0 \leq θ \leq 2 π

, then

θ = - - - -

Join BYJU'S Learning Program

If sin3θ=sinθ, how many solutions exist such that −2π<θ<2π ?

The correct option is C 11We have, sin3θ=sinθ ⇒sin3θ−sinθ=0 ⇒2cos(3θ+θ2)sin(3θ−θ2)=0 ⇒cos2θ.sinθ=0 ⇒cos2θ=0 or sinθ=0 ⇒θ=nπ±π4 or θ=nπ,n∈I ∴θ=±π4,±3π4,±5π4,±7π4,0,±π (∵−2π <θ<2π) Thus, total number of solutions = 11

If $sin 3 θ = sin θ,$ how many solutions exist such that $- 2 π < θ < 2 π$ ?