The true set of values of k for which sin -1((1/1+ sin 2 x))=(k π/6) may have a solution, is

Question

The true set of values of k for which sin -1((1/1+ sin 2 x))=(k π/6) may have a solution, is (A) [(π/4), (π/2)] (B) [1,3] (C) [(1/6), (1/2)] (D) [2,4]

Tardigrade Admin · Accepted Answer

As, 1 ≤ 1+ sin 2 x ≤ 2 ⇒ (1/2) ≤ (1/1+ sin 2 x) ≤ 1 ⇒ (π/6) ≤ sin -1((1/1+ sin 2 x)) ≤ (π/2) ⇒ (π/6) ≤ ( k π/6) ≤ (π/2) ⇒ 1 ≤ k ≤ 3

Q. The true set of values of $k$ for which $sin^{- 1} (\frac{1}{1 + s i n ^{2} x}) = \frac{kπ}{6}$ may have a solution, is

$[\frac{π}{4}, \frac{π}{2}]$

$[1, 3]$

$[\frac{1}{6}, \frac{1}{2}]$

$[2, 4]$

As, $1 \leq 1 + sin^{2} x \leq 2 \Rightarrow \frac{1}{2} \leq \frac{1}{1 + s i n ^{2} x} \leq 1$
$\Rightarrow \frac{π}{6} \leq sin^{- 1} (\frac{1}{1 + s i n ^{2} x}) \leq \frac{π}{2}$
$\Rightarrow \frac{π}{6} \leq \frac{kπ}{6} \leq \frac{π}{2} \Rightarrow 1 \leq k \leq 3$

Q. The true set of values of k for which sin−1(1+sin2x1​)=6kπ​ may have a solution, is

[4π​,2π​]

[1,3]

[61​,21​]

[2,4]