Question

Solve for x, the given inverse trigonometric expression ${{\cos }^{-1}}x+{{\sin }^{-1}}\dfrac{x}{2}=\dfrac{\pi }{6}$.

Answer 1

In this question, we have two different inverses trigonometric functions. So, we need to make our LHS have one inverse trigonometric function. For that, we just need to shift the inverse sine function to RHS and then our LHS will have only inverse cosine function.
Then, take cosine in both LHS and RHS and use the formula$\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$. We consider $\theta ={{\sin }^{-1}}\dfrac{x}{2}$ . Transform $\sin \theta$ into $\cos \theta$ and then solve it further.

Complete step-by-step solution:
According to the question, we have ${{\cos }^{-1}}x+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)=\dfrac{\pi }{6}$………….(1)
Now, shifting the inverse sine function to RHS in equation (1)
$\Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{6}-{{\sin }^{-1}}\left( \dfrac{x}{2} \right)$…………..(2)
Now, taking cosine in LHS as well as RHS in equation (2), we get
$\Rightarrow x=\cos \left( \dfrac{\pi }{6}-{{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)$…………..(3)
Expand the equation (3), using the formula $\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$.
After expanding we get,
$\Rightarrow x=\cos \dfrac{\pi }{6}\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)+\sin \dfrac{\pi }{6}\sin \left( {{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)$…………..(4)
We know the property, $\sin \left( {{\sin }^{-1}}(x) \right)=x$ .
Replacing x by $\dfrac{x}{2}$ in the above property, we get $\sin \left( {{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)=\dfrac{x}{2}$ …………………..(5)
Now, using equation (5) and putting the values of $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ and $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ in equation (4), we get
$\Rightarrow x=\dfrac{\sqrt{3}}{2}\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)+\dfrac{1}{2}.\dfrac{x}{2}$ ………(6)
In equation (6), we have an inverse trigonometric function. So, we have to simplify that inverse function.
Here let us assume,
$\theta ={{\sin }^{-1}}\dfrac{x}{2}$ …………(7)
$\Rightarrow \sin \theta =\dfrac{x}{2}$………….(8)
We know the identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Shifting, ${{\sin }^{2}}\theta$ to the RHS, we get
$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$ …………….(9)
Substituting equation (8), in equation (9), we get

& \cos \theta =\sqrt{1-{{\sin }^{2}}\theta } \\\ & \Rightarrow \cos \theta =\sqrt{1-\dfrac{{{x}^{2}}}{4}} \\\ \end{aligned}$$ The above equation can also be written as, $$\theta ={{\cos }^{-1}}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right)$$ …………….(8) Here the term under square root should be greater than or equal to 0. That is, $$1-\dfrac{{{x}^{2}}}{4}\ge 0$$ . From equation (7) we have, $$\theta ={{\sin }^{-1}}\dfrac{x}{2}$$. Also from equation (9), we have $$\theta ={{\cos }^{-1}}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right)$$. Equation (7) and equation (9) are equal. So, $$\theta ={{\sin }^{-1}}\dfrac{x}{2}={{\cos }^{-1}}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right)$$ ………..(10) From equation (6), we have $$\Rightarrow x=\dfrac{\sqrt{3}}{2}\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{2} \right) \right)+\dfrac{1}{2}.\dfrac{x}{2}$$ From equation (9), we can write equation (5) as written below, $$x=\dfrac{\sqrt{3}}{2}\cos \left( {{\cos }^{-1}}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right) \right)+\dfrac{x}{4}$$ ………………….(11) We know the property, $$\cos \left( {{\cos }^{-1}}(x) \right)=x$$ . Replacing x by $$\sqrt{1-\dfrac{{{x}^{2}}}{4}}$$ in the above property, we get $$\cos \left( {{\cos }^{-1}}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right) \right)=\sqrt{1-\dfrac{{{x}^{2}}}{4}}$$ ……………….(12) From equation (11) and equation (12), we get $$\begin{aligned} & \Rightarrow x-\dfrac{x}{4}=\dfrac{\sqrt{3}}{2}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right) \\\ & \Rightarrow \dfrac{3x}{4}=\dfrac{\sqrt{3}}{2}\left( \sqrt{1-\dfrac{{{x}^{2}}}{4}} \right) \\\ & \Rightarrow \dfrac{9{{x}^{2}}}{16}=\dfrac{3}{4}\left( 1-\dfrac{{{x}^{2}}}{4} \right) \\\ & \Rightarrow 3{{x}^{2}}=4-{{x}^{2}} \\\ & \Rightarrow 4{{x}^{2}}=4 \\\ & \Rightarrow {{x}^{2}}=1 \\\ & \Rightarrow x=\pm 1 \\\ \end{aligned}$$ We take only x=1, not x=-1. Because x=-1 doesn’t satisfy the equation $${{\cos }^{-1}}x+{{\sin }^{-1}}\dfrac{x}{2}=\dfrac{\pi }{6}$$ . **Hence, x=1 is the solution.** **Note:** In this question, one might do a silly mistake while taking the value of x. After solving, we get two values of x that are 1 and -1. Here, one might also take $$x=-1$$ and conclude it as an answer. This is wrong because when we put the value of x as -1 in the equation $${{\cos }^{-1}}x+{{\sin }^{-1}}\dfrac{x}{2}=\dfrac{\pi }{6}$$ , we get $$\dfrac{5\pi }{6}$$ as LHS and in RHS, we have $$\dfrac{\pi }{6}$$ . So, we can say that the LHS is not equal to the RHS. Hence, $$x=-1$$ doesn’t satisfies the given equation, $${{\cos }^{-1}}x+{{\sin }^{-1}}\dfrac{x}{2}=\dfrac{\pi }{6}$$ . Therefore, we must verify the values of x after putting it in the given equation and then conclude the answer.