Question

The value of \(x\) that satisfies the equation \[ \sin^{-1}(x)=\cos^{-1}\!\left(\frac{3x}{4}\right) \] is: 

• A. \(\frac{4}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{4}{9}\)
• D. \(\frac{3}{9}\)
• E. \(\frac{2}{5}\)

Hint:

Always convert all inverse trigonometric terms to the same type (all sin-inverse or all cos-inverse). This allows you to set the arguments equal to each other.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To solve an equation with different inverse trigonometric functions, we can represent them using the same function, typically by treating the arguments as sides of a right-angled triangle.

Step 2: Key Formula or Approach:
Let \(\sin^{-1}(x) = \theta \implies \sin \theta = x\).
From a right triangle, \(\cos \theta = \sqrt{1 - x^2}\).
The equation becomes \(\cos^{-1}(\sqrt{1 - x^2}) = \cos^{-1}(\frac{3x}{4})\).

Step 3: Detailed Explanation:
By removing the \(\cos^{-1}\) from both sides:
\[ \sqrt{1 - x^2} = \frac{3x}{4} \]
Square both sides:
\[ 1 - x^2 = \frac{9x^2}{16} \]
\[ 1 = x^2 + \frac{9x^2}{16} \]
\[ 1 = \frac{16x^2 + 9x^2}{16} \]
\[ 1 = \frac{25x^2}{16} \]
\[ x^2 = \frac{16}{25} \]
Taking the positive square root (since inverse sine of negative would not match inverse cosine of negative here):
\[ x = \frac{4}{5} \]

Step 4: Final Answer:
The value of \(x\) is \(\frac{4}{5}\).