Number of solutions of the equation sin -11-x-2 sin -1 x-2 isA- 1B- 0C- 2D- -

The correct option is A 1Let x = sin y ⇒ sin^ 1(1 sin y) 2 sin^ 1(sin y)=π/2 ⇒ sin^ 1(1 siny)=π/2+2y ⇒ 1 sin y=cos 2y ⇒ 1 cos 2y=sin y ⇒ 2 sin^2y sin y = 0 ...

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Byju's Answer

Standard XII

Mathematics

Exponential Function (a^x)

Number of sol...

Question

Number of solutions of the equation $s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}$ is

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Solution

The correct option is A 1
Let x = sin y
$\Rightarrow s i n^{- 1} (1 - s i n y) - 2 s i n^{- 1} (s i n y) = \frac{π}{2}$
$\Rightarrow s i n^{- 1} (1 - s i n y) = \frac{π}{2} + 2 y$
$\Rightarrow 1 - s i n y = c o s 2 y$
$\Rightarrow 1 - c o s 2 y = s i n y \Rightarrow 2 s i n^{2} y - s i n y = 0$
$\Rightarrow s i n y = 0 (o r) s i n y = \frac{1}{2}$
$\Rightarrow x = 0 (o r) x = \frac{1}{2}$
$B u t x = \frac{1}{2}$ is pseudo result
$∴$ x = 0 is the only solution

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Number of solutions of the equation sin–1(1−x)−2sin–1x=π2 is

The correct option is A 1Let x = sin y ⇒sin−1(1−sin y)−2 sin−1(sin y)=π2 ⇒sin−1(1−siny)=π2+2y ⇒1−sin y=cos 2y ⇒1−cos 2y=sin y⇒2 sin2y−sin y=0 ⇒sin y=0 (or) sin y=12 ⇒x=0 (or) x=12 But x=12 is pseudo result ∴ x = 0 is the only solution

Number of solutions of the equation $s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}$ is