Question

Sin inverse x + sin inverse 2x=pi upon 3

Answer 1

sin^-1(x)+sin^-1(2x)=π/3

sin^-1(2x)=π/3-sin^-1(x)

2x=sin(π/3-sin^-1(x))

2x=sin(π/3)cos(sin^−1(x))−cos(π/3)sin(sin^−1(x))

2x=√3/2cos(cos^−1(whole root over 1−x^2))−1/2(x)

2x+x/2=√3/2(whole root over 1-x²)

5x=√3(whole root over 1-x²)

25x²=3-3x²

28x²=3

x=√(3/28)

Answer 2

Answer:

Required result  $x=\sqrt{\frac{3}{28}}$

Step-by-step explanation:

Given : $\sin^{-1}(x)+\sin^{-1}(2x)=\frac{\pi}{3}$

To find : Solve for x?

Solution :

The given expression is

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

Now, we solve the expression for x,

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

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

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

$2x=\frac{\sqrt3}{2}\cos(\cos^{-1}\sqrt{1-x^2})-\frac12(x)$

$2x+\frac x2=\frac{\sqrt3}{2}(\sqrt{1-x^2})$

$5x=\sqrt3(\sqrt{1-x^2})$

$(5x)^2=(\sqrt3(\sqrt{1-x^2}))^2$

$25x^2=3-3x^2$

 $28x^2=3$

$x=\pm\sqrt{\frac{3}{28}}$

But, only positive value of x satisfy the original equation,

Therefore, $x=\sqrt{\frac{3}{28}}$