If sin θ, sin 2 θ, 1 are the first 3 terms of an A.P., then the number of values of θ in (-π, π), is

Question

If sin θ, sin 2 θ, 1 are the first 3 terms of an A.P., then the number of values of θ in (-π, π), is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

sin θ, sin 2 θ, 1 arrow text A.P. 2 sin 2 θ=1+ sin θ sin θ=1 text or sin θ=-(1/2) θ=(π/2) text or θ=(-π/6) text or θ=(-5 π/6) Hence number of values of θ are 3 .

Q. If $sin θ, sin^{2} θ, 1$ are the first 3 terms of an A.P., then the number of values of $θ$ in $(- π, π)$ , is

1

2

3

4

$sin θ, sin^{2} θ, 1 \to A.P.$
$2 sin^{2} θ = 1 + sin θ$
$sin θ = 1 or sin θ = - \frac{1}{2}$
$θ = \frac{π}{2} or θ = \frac{- π}{6} or θ = \frac{- 5 π}{6} < b r / >$
Hence number of values of $θ$ are 3 .

Q. If sinθ,sin2θ,1 are the first 3 terms of an A.P., then the number of values of θ in (−π,π), is

1

2

3

4

sinθ,sin2θ,1→ A.P. 2sin2θ=1+sinθ sinθ=1 or sinθ=−21​ θ=2π​ or θ=6−π​ or θ=6−5π​<br/> Hence number of values of θ are 3 .

Q. If $sin θ, sin^{2} θ, 1$ are the first 3 terms of an A.P., then the number of values of $θ$ in $(- π, π)$ , is

$sin θ, sin^{2} θ, 1 \to A.P.$
$2 sin^{2} θ = 1 + sin θ$
$sin θ = 1 or sin θ = - \frac{1}{2}$
$θ = \frac{π}{2} or θ = \frac{- π}{6} or θ = \frac{- 5 π}{6} < b r / >$
Hence number of values of $θ$ are 3 .