Problem
Solve: \( (\sin^{-1}x)^2 + (\cos^{-1}x)^2 = \frac{17\pi^2}{36} \)
Solution
Step 1: Use identity
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Let \( a = \sin^{-1}x \), then:
\[ \cos^{-1}x = \frac{\pi}{2} – a \]
Step 2: Substitute
\[ a^2 + \left(\frac{\pi}{2} – a\right)^2 = \frac{17\pi^2}{36} \]
Step 3: Expand
\[ a^2 + \frac{\pi^2}{4} – \pi a + a^2 = \frac{17\pi^2}{36} \]
\[ 2a^2 – \pi a + \frac{\pi^2}{4} = \frac{17\pi^2}{36} \]
Step 4: Simplify
\[ 2a^2 – \pi a + \left(\frac{9\pi^2}{36} – \frac{17\pi^2}{36}\right) = 0 \]
\[ 2a^2 – \pi a – \frac{8\pi^2}{36} = 0 \]
\[ 2a^2 – \pi a – \frac{2\pi^2}{9} = 0 \]
Step 5: Solve quadratic
\[ a = \frac{\pi \pm \sqrt{\pi^2 + \frac{16\pi^2}{9}}}{4} \]
\[ = \frac{\pi \pm \frac{5\pi}{3}}{4} \]
So,
\[ a = \frac{2\pi}{3} \quad \text{or} \quad a = -\frac{\pi}{6} \]
Step 6: Valid solution
Since \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \),
\[ a = -\frac{\pi}{6} \]
Step 7: Find x
\[ x = \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \]
Final Answer
\[ \boxed{-\frac{1}{2}} \]
Explanation
Use identity sin⁻¹x + cos⁻¹x = π/2 and reduce the equation to a quadratic.