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Find the value of $x$, if $\sin^{-1}{x}+\sin^{-1}{2x}=\dfrac{\pi}{3}$

$x$ is a value that represents the ratio of length of the opposite side to length of the hypotenuse of a triangle. $2x$ is another value that represents the ratio of length of opposite side to length of the hypotenuse of another triangle. The sum of inverse sines of both values is equal to $60^\circ$ and it is expressed in mathematical form.


The inverse trigonometric equation can be solved by applying the sum rule of inverse sine functions to obtain the value of $x$ but it makes the simplification more complicated in this case.


Technique to simplify the equation

According to trigonometry, the value of sine of angle $\dfrac{\pi}{3}$ is $\dfrac{\sqrt{3}}{2}$. So, the angle $\dfrac{\pi}{3}$ can be expressed as inverse sine of $\dfrac{\sqrt{3}}{2}$ as per inverse trigonometry.

$\implies \sin^{-1}{x}+\sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}$

Shift any inverse sine function to right hand side expression of this equation and then start simplification to get the value of $x$ mathematically.

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}-\sin^{-1}{x}$


Combining the inverse trigonometric functions

The right hand side expression represents the subtraction of an inverse sine function from another inverse sine function but it can be combined by applying subtraction rule of inverse sine functions.

$\sin^{-1}{x}-\sin^{-1}{y}$ $\,=\,$ $\sin^{-1}{\Big[x\sqrt{1-y^2}-y\sqrt{1-x^2}\Big]}$

According to this formula, take $x = \dfrac{\sqrt{3}}{2}$and $y = x$.

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{1-{\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}^2}\Bigg]}$

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{1-\dfrac{3}{4}}\Bigg]}$

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{\dfrac{4-3}{4}}\Bigg]}$

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{\dfrac{1}{4}}\Bigg]}$

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-x \times \dfrac{1}{2}\Bigg]}$

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}-\dfrac{x}{2}\Bigg]}$


Eliminating Inverse Trigonometric functions

$\implies \sin^{-1}{2x}$ $\,=\,$ $\sin^{-1}{\Bigg[\dfrac{\sqrt{3}\sqrt{1-x^2}-x}{2}\Bigg]}$

Observe this inverse trigonometric equation. The both sides contains same inverse trigonometric functions. Hence, the value inside the inverse sine function of left hand side expression is equal to the value inside the inverse sine function of right hand side expression. Therefore, eliminate the inverse sines from both sides.

$\implies 2x$ $\,=\,$ $\dfrac{\sqrt{3}\sqrt{1-x^2}-x}{2}$


Solving the value of x

Use cross multiplication technique and proceed to simplify the equation for the value of $x$.

$\implies 4x$ $\,=\,$ $\sqrt{3}\sqrt{1-x^2}-x$

$\implies 4x+x$ $\,=\,$ $\sqrt{3}\sqrt{1-x^2}$

$\implies 5x$ $\,=\,$ $\sqrt{3}\sqrt{1-x^2}$

Take square both sides to eliminate the square root from the right hand side expression.

$\implies {(5x)}^2$ $\,=\,$ ${(\sqrt{3}\sqrt{1-x^2})}^2$

$\implies 25x^2$ $\,=\,$ ${(\sqrt{3})}^2 {(\sqrt{1-x^2})}^2$

$\implies 25x^2$ $\,=\,$ $3(1-x^2)$

$\implies 25x^2$ $\,=\,$ $3-3x^2$

$\implies 25x^2+3x^2 \,=\, 3$

$\implies 28x^2 \,=\, 3$

$\implies x^2 \,=\, \dfrac{3}{28}$

$\implies x \,=\, \pm \sqrt{\dfrac{3}{28}}$

$\implies x \,=\, \pm \sqrt{\dfrac{3}{4 \times 7}}$

$\implies x \,=\, \pm \dfrac{\sqrt{3}}{2\sqrt{7}}$

Therefore, $x \,=\, \dfrac{\sqrt{3}}{2\sqrt{7}}$ and $x \,=\, -\dfrac{\sqrt{3}}{2\sqrt{7}}$.


Verifying the value of x

Consider the actual inverse trigonometric equation to determine the value of $x$ logically.


The equation has given two solutions and they are $x \,=\, \dfrac{\sqrt{3}}{2\sqrt{7}}$ and $x \,=\, -\dfrac{\sqrt{3}}{2\sqrt{7}}$

If $x \,=\, \dfrac{\sqrt{3}}{2\sqrt{7}}$, then the value of $\sin^{-1}{\Bigg(\dfrac{\sqrt{3}}{2\sqrt{7}}\Bigg)}$ and $\sin^{-1}{\Bigg(\dfrac{2\sqrt{3}}{2\sqrt{7}}\Bigg)}$ are positive. So, it is possible to obtain a positive value and it is $\dfrac{\pi}{3}$.

But if $x \,=\, -\dfrac{\sqrt{3}}{2\sqrt{7}}$, then the value of $\sin^{-1}{\Bigg(-\dfrac{\sqrt{3}}{2\sqrt{7}}\Bigg)}$ and $\sin^{-1}{\Bigg(-\dfrac{2\sqrt{3}}{2\sqrt{7}}\Bigg)}$ are negative. They both give negative angles. Therefore, it is not possible to obtain a positive value $\dfrac{\pi}{3}$ by sum of them.

Therefore, $x \,=\, \dfrac{\sqrt{3}}{2\sqrt{7}}$ is the true solution and $x \,\ne\, \dfrac{-\sqrt{3}}{2\sqrt{7}}$. It is required solution for this inverse trigonometry math problem.

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