Question

Given that the inverse trigonometric functions take principal values only. Find the number of real values of \[x\] which satisfy \[{\sin ^{ - 1}}\left( {\dfrac{{3x}}{5}} \right) + {\sin ^{ - 1}}\left( {\dfrac{{4x}}{5}} \right) = {\sin ^{ - 1}}x\].
A. 1
B. 2
C. 3
D. 0

Answer 1

Hint: First we will apply the \[{\sin ^{ - 1}}a + {\sin ^{ - 1}}b\] formula to the left side and apply an sine function to remove \[{\sin ^{ - 1}}\] from both sides. Then solve the equation to find the number of solutions of \[x\].

Formula used:
\[{\sin ^{ - 1}}a + {\sin ^{ - 1}}b = {\sin ^{ - 1}}\left( {a\sqrt {1 - {a^2}} + b\sqrt {1 - {a^2}} } \right)\]
\[\sin \left( {{{\sin }^{ - 1}}x} \right) = x\]

Complete step by step solution:
Given expression is
\[{\sin ^{ - 1}}\left( {\dfrac{{3x}}{5}} \right) + {\sin ^{ - 1}}\left( {\dfrac{{4x}}{5}} \right) = {\sin ^{ - 1}}x\]
Now apply the sum of \[{\sin ^{ - 1}}\].
\[ \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{{3x}}{5}\sqrt {1 - {{\left( {\dfrac{{4x}}{5}} \right)}^2}} + \dfrac{{4x}}{5}\sqrt {1 - {{\left( {\dfrac{{3x}}{5}} \right)}^2}} } \right) = {\sin ^{ - 1}}x\]
Simplify the above equation
\[ \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{{3x}}{5}\sqrt {1 - \dfrac{{16{x^2}}}{{25}}} + \dfrac{{4x}}{5}\sqrt {1 - \dfrac{{9{x^2}}}{{25}}} } \right) = {\sin ^{ - 1}}x\]
Now take \[\sin \] both sides
\[ \Rightarrow \sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{{3x}}{5}\sqrt {1 - \dfrac{{16{x^2}}}{{25}}} + \dfrac{{4x}}{5}\sqrt {1 - \dfrac{{9{x^2}}}{{25}}} } \right)} \right] = \sin \left[ {{{\sin }^{ - 1}}x} \right]\]
Apply the formula \[\sin \left( {{{\sin }^{ - 1}}x} \right) = x\]
\[ \Rightarrow \dfrac{{3x}}{5}\sqrt {1 - \dfrac{{16{x^2}}}{{25}}} + \dfrac{{4x}}{5}\sqrt {1 - \dfrac{{9{x^2}}}{{25}}} = x\]
Simplify the expression under the square root.
\[ \Rightarrow \dfrac{{3x}}{5}\sqrt {\dfrac{{25 - 16{x^2}}}{{25}}} + \dfrac{{4x}}{5}\sqrt {\dfrac{{25 - 9{x^2}}}{{25}}} = x\]
Subtract \[x\] from both sides of equation
\[ \Rightarrow \dfrac{{3x}}{5}\sqrt {\dfrac{{25 - 16{x^2}}}{{25}}} + \dfrac{{4x}}{5}\sqrt {\dfrac{{25 - 9{x^2}}}{{25}}} - x = 0\]
Take common \[x\]
\[ \Rightarrow x\left( {\dfrac{3}{5}\sqrt {\dfrac{{25 - 16{x^2}}}{{25}}} + \dfrac{4}{5}\sqrt {\dfrac{{25 - 9{x^2}}}{{25}}} - 1} \right) = 0\]
Equate each factor by zero
Either \[x = 0\]
Or, \[\dfrac{3}{5}\sqrt {\dfrac{{25 - 16{x^2}}}{{25}}} + \dfrac{4}{5}\sqrt {\dfrac{{25 - 9{x^2}}}{{25}}} - 1 = 0\]
Rewrite the above equation
\[ \Rightarrow \dfrac{3}{5}\sqrt {\dfrac{{25 - 16{x^2}}}{{25}}} = 1 - \dfrac{4}{5}\sqrt {\dfrac{{25 - 9{x^2}}}{{25}}} \]
\[ \Rightarrow \dfrac{3}{{25}}\sqrt {25 - 16{x^2}} = 1 - \dfrac{4}{{25}}\sqrt {25 - 9{x^2}} \]
\[ \Rightarrow \dfrac{3}{{25}}\sqrt {25 - 16{x^2}} = \dfrac{{25 - 4\sqrt {25 - 9{x^2}} }}{{25}}\]
Cancel out of \[\dfrac{1}{{25}}\] from both sides
\[ \Rightarrow 3\sqrt {25 - 16{x^2}} = 25 - 4\sqrt {25 - 9{x^2}} \]
Raise power 2 both sides
\[ \Rightarrow {\left( {3\sqrt {25 - 16{x^2}} } \right)^2} = {\left( {25 - 4\sqrt {25 - 9{x^2}} } \right)^2}\]
Apply algebraical identity to simplify
\[ \Rightarrow 9\left( {25 - 16{x^2}} \right) = {25^2} + {\left( {4\sqrt {25 - 9{x^2}} } \right)^2} - 2 \cdot 25 \cdot \left( {4\sqrt {25 - 9{x^2}} } \right)\]
\[ \Rightarrow 225 - 144{x^2} = 625 + 16 \cdot \left( {25 - 9{x^2}} \right) - 200\sqrt {25 - 9{x^2}} \]
\[ \Rightarrow 225 - 144{x^2} = 625 + 400 - 144{x^2} - 200\sqrt {25 - 9{x^2}} \]
\[ \Rightarrow 225 - 144{x^2} = 1025 - 144{x^2} - 200\sqrt {25 - 9{x^2}} \]
Rewrite the equation such that all radical part should be one side
\[ \Rightarrow 225 - 144{x^2} - 1025 + 144{x^2} = - 200\sqrt {25 - 9{x^2}} \]
\[ \Rightarrow - 800 = - 200\sqrt {25 - 9{x^2}} \]
Divide both sides -200
\[ \Rightarrow 4 = \sqrt {25 - 9{x^2}} \]
Take square both sides of the equation
 \[ \Rightarrow {4^2} = {\left( {\sqrt {25 - 9{x^2}} } \right)^2}\]
\[ \Rightarrow 16 = 25 - 9{x^2}\]
Subtract 25 from both sides
\[ \Rightarrow 16 - 25 = 25 - 9{x^2} - 25\]
\[ \Rightarrow - 9 = - 9{x^2}\]
Divide both sides by -9
\[ \Rightarrow 1 = {x^2}\]
\[ \Rightarrow x = \pm 1\]
The solution of \[x\] are 0, 1, -1.
The number of solutions of \[x\] are 3.
Hence option C is the correct option.

Note: Students often confused \[{\sin ^{ - 1}}x\] with \[{\left( {\sin x} \right)^{ - 1}}\]. In fact \[{\left( {\sin x} \right)^{ - 1}} = \dfrac{1}{{\sin x}}\] and similar to other function. But \[{\sin ^{ - 1}}x\] is a function.
Inverse of a function can be written as \[\arcsin \], \[\arccos \], \[\arctan \] etc.