Question

Find the value of x if,
${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \dfrac{\pi }{6} = 0$.

Answer 1

Here first we will simplify the given equation and then apply the following formula:
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Then solve for the value of x.

Complete step-by-step answer:
The given equation is:-
${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \dfrac{\pi }{6} = 0$
Simplifying the equation we get:-

$${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{\pi }{6} \\\ \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{6} - {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) \\\$$

Now taking cos of both the sides we get:-
$\cos \left( {{{\cos }^{ - 1}}x} \right) = \cos \left( {\dfrac{\pi }{6} - {{\sin }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)$
Now we know that:
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\cos \left( {{{\cos }^{ - 1}}x} \right) = x$
Hence applying the formulas we get:-
$x = \cos \left( {\dfrac{\pi }{6}} \right)\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) + \sin \left( {\dfrac{\pi }{6}} \right)\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)$
Now we know that:-

$$\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2} \\\ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} \\\ \sin \left( {{{\sin }^{ - 1}}x} \right) = x \\\$$

Putting these values we get:-
$x = \dfrac{{\sqrt 3 }}{2}\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) + \dfrac{1}{2} \times \dfrac{x}{2}$……………………………(1)
Now since,
${\sin ^{ - 1}}\theta = {\cos ^{ - 1}}\left( {\sqrt {1 - {\theta ^2}} } \right)$
Hence applying this formula we get:-
${\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = {\cos ^{ - 1}}\left( {\sqrt {1 - {{\left( {\dfrac{x}{2}} \right)}^2}} } \right)$
Solving it further we get:-

$${\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = {\cos ^{ - 1}}\left( {\sqrt {\dfrac{{4 - {x^2}}}{4}} } \right) \\\ \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt {4 - {x^2}} }}{2}} \right) \\\$$

Hence putting this value in equation1 we get:-
$x = \dfrac{{\sqrt 3 }}{2}\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{{\sqrt {4 - {x^2}} }}{2}} \right)} \right) + \dfrac{1}{2} \times \dfrac{x}{2}$
Since we know that:
$\cos \left( {{{\cos }^{ - 1}}x} \right) = x$
Hence applying this formula we get:-
$x = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt {4 - {x^2}} }}{2} + \dfrac{x}{4}$
Solving it further we get:-

$$\Rightarrow x = \dfrac{{\sqrt {3\left( {4 - {x^2}} \right)} }}{4} + \dfrac{x}{4} \\\ \Rightarrow x - \dfrac{x}{4} = \dfrac{{\sqrt {3\left( {4 - {x^2}} \right)} }}{4} \\\ \Rightarrow \dfrac{{4x - x}}{4} = \dfrac{{\sqrt {3\left( {4 - {x^2}} \right)} }}{4} \\\ \Rightarrow \dfrac{{3x}}{4} = \dfrac{{\sqrt {3\left( {4 - {x^2}} \right)} }}{4} \\\$$

Simplifying it further we get:-
$3x = \sqrt {3\left( {4 - {x^2}} \right)}$
Now squaring both the sides we get:-

$${\left( {3x} \right)^2} = {\left( {\sqrt {3\left( {4 - {x^2}} \right)} } \right)^2} \\\ \Rightarrow 9{x^2} = 3\left( {4 - {x^2}} \right) \\\$$

Now simplifying it further we get:-

$$9{x^2} = 12 - 3{x^2} \\\ 9{x^2} + 3{x^2} = 12 \\\ 12{x^2} = 12 \\\$$

Dividing both sides by 12 we get:-
${x^2} = 1$
Now taking square root of both the sides we get:-

$$\sqrt {{x^2}} = \sqrt 1 \\\ x = \pm 1 \\\$$

But the value of x cannot be -1
Hence we get:-
$x = 1$

Note: Students should note that in such questions we have converted the question into one form either of cosine or sine and then apply the known identities to solve it.
Also while taking the square root of any quantity both the positive and negative values should be considered.