Question

Sin^-1 x+ sin^-2x=π/3​

Answer 1

Answer:

⇒ $x=\sqrt{\frac{3}{28}}$

Step-by-step explanation:

Given :

Find x, if

$sin^{-1}x+sin^{-1}2x=\frac{\pi}{3}$

Solution :

$sin^{-1}x+sin^{-1}2x=\frac{\pi}{3}$

$sin^{-1}2x=\frac{\pi}{3} -sin^{-1}x$

By taking sin functions, both the sides,.

$2x= sin(\frac{\pi}{3} - sin^{-1}x)$ (here, $sin^{-1}x)$ is treated as an angle θ)

By using ,

sin (A - B) = sin A cos B - cos A sin B

⇒ $2x=sin\frac{\pi}{3}cos(sin^{-1}x) - cos\frac{\pi}{3}sin(sin^{-1}x)$

⇒ $2x=\frac{\sqrt{3}}{2}[\sqrt{ 1- sin^2(sin^{-1}x)}] - \frac{1}{2}(x)$

⇒ $2x=\frac{\sqrt{3}}{2}[\sqrt{ 1- x^2}] - \frac{x}{2}$

⇒ $2x+ \frac{x}{2}=\frac{\sqrt{3}}{2}[\sqrt{ 1- x^2}]$

⇒ $\frac{5x}{2}=\frac{\sqrt{3}}{2}[\sqrt{1- x^2}]$

⇒ $5x=\sqrt{3}[\sqrt{ 1- x^2}]$

By squaring both the sides,

We get,

⇒ $(5x)^2=(\sqrt{3}[\sqrt{ 1- x^2}])^2$

⇒ $25x^2 = 3(1-x^2)$

⇒ $25x^2 = 3-3x^2$

⇒ $25x^2+3x^2= 3$

⇒ $28x^2= 3$

⇒ $x^2=\frac{3}{28}$

⇒ $x=$±$\sqrt{\frac{3}{28}}$

But, Only positive value of x satisfy the given equation,

Hence,

⇒ $x=\sqrt{\frac{3}{28}}$