The interval in which - belongs- such that the inequality 2 sin 2-3-sin-3-1 - 0 is satisfied and - - if

The correct options are A [ π6, π2 ] C [ π2, 5 π6 ]Let sin ( θ π3 )=x ∴ 2x^2 x 1 ≤ 0 ⇒ (2x+1)(x 1) ≤ 0 ⇒ x ∈ [ 12, 1 ] 12≤sin ( θ π3 )≤ 1 Let θ π3=α Th ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

The interval ...

Question

The interval in which $θ$ belongs, such that the inequality $2 {sin}^{2} (θ - \frac{π}{3}) - sin (θ - \frac{π}{3}) - 1 \leq 0$ is satisfied and $θ \in [- π, π]$ is

[\frac{π}{6}, \frac{π}{2}]

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[- \frac{π}{6}, \frac{π}{6}]

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[\frac{π}{2}, \frac{5 π}{6}]

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[- \frac{5 π}{6}, \frac{5 π}{6}]

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Solution

The correct options are
A $[\frac{π}{6}, \frac{π}{2}]$
C $[\frac{π}{2}, \frac{5 π}{6}]$
Let $sin (θ - \frac{π}{3}) = x$
$∴ 2 x^{2} - x - 1 \leq 0$
$\Rightarrow (2 x + 1) (x - 1) \leq 0$
$\Rightarrow x \in [- \frac{1}{2}, 1]$
$- \frac{1}{2} \leq sin (θ - \frac{π}{3}) \leq 1$

Let $θ - \frac{π}{3} = α$
Then $- \frac{1}{2} \leq sin α \leq 1$

From graph,
$\frac{- π}{6} \leq θ - \frac{π}{3} \leq π$
$\Rightarrow \frac{π}{6} \leq θ \leq π as θ_{max} = π$

$\frac{- 3 π}{2} \leq θ - \frac{π}{3} \leq \frac{- 5 π}{6}$
$\Rightarrow \frac{- 7 π}{6} \leq θ \leq \frac{- π}{2}$
$\Rightarrow - π \leq θ \leq \frac{- π}{2} as θ_{min} = - π$

$∴ θ \in [- π, \frac{- π}{2}] \cup [\frac{π}{6}, π]$

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The interval in which θ belongs, such that the inequality 2sin2(θ−π3)−sin(θ−π3)−1≤0 is satisfied and θ∈[−π,π] is

The interval in which $θ$ belongs, such that the inequality $2 {sin}^{2} (θ - \frac{π}{3}) - sin (θ - \frac{π}{3}) - 1 \leq 0$ is satisfied and $θ \in [- π, π]$ is