Question & Answer
QUESTION

Find the value of x for the given inverse trigonometric function $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi $.

ANSWER Verified Verified
Hint: In order to solve the problem first try to convert these two trigonometric functions into one by the use of different inverse trigonometric identities and then move on with the simplification part and calculate the value of x.

Complete step-by-step answer:
Given equation is: $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi $
We need to find x from this equation,
As we know that
$
  {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} \\
   \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\
 $
Let us substitute the value from above formula into the given equation
\[
   \Rightarrow 4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi \\
   \Rightarrow 4{\sin ^{ - 1}}x + \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right) = \pi \\
 \]
Now let us simplify the equation to find the value of x
\[
   \Rightarrow 3{\sin ^{ - 1}}x + \dfrac{\pi }{2} = \pi \\
   \Rightarrow 3{\sin ^{ - 1}}x = \pi - \dfrac{\pi }{2} \\
   \Rightarrow 3{\sin ^{ - 1}}x = \dfrac{\pi }{2} \\
   \Rightarrow {\sin ^{ - 1}}x = \dfrac{1}{3} \times \dfrac{\pi }{2} \\
   \Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{6} \\
 \]
Now, let us bring the \[{\sin ^{ - 1}}\] function from the LHS to the RHS. So it will convert to $\sin $ function.
$
   \Rightarrow x = \sin \left( {\dfrac{\pi }{6}} \right) \\
   \Rightarrow x = \sin \left( {\dfrac{{{{180}^0}}}{6}} \right){\text{ }}\left[ {\because \pi = {{180}^0}} \right] \\
   \Rightarrow x = \sin \left( {{{30}^0}} \right) \\
 $
As we know the value of $\sin {30^0}$ from the trigonometric table. So we will directly substitute the value.
$
  x = \sin {30^0} = \dfrac{1}{2} \\
   \Rightarrow x = \dfrac{1}{2} \\
 $
Hence, the value of x is $\dfrac{1}{2}$ .

Note: In order to solve such problems students must remember the formulas for the inverse trigonometry. Also students must learn the values of the trigonometric terms for some common angles. This question would have been impossible to solve if we would not have converted the two functions into one so students must recognize which functions to remove.