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Inverse Trigonometric Functions: Number Points Satisfying Equation

JEE Maths question with a full step-by-step solution.

Question
The number of points x∈[−π2,3π2]x\in\left[-\dfrac{\pi}{2},\dfrac{3\pi}{2}\right] satisfying the equation 1+sin⁡−1(sin⁡x)=π31+\sin^{-1}(\sin x)=\dfrac{\pi}{3}.
Solution
Answer: 2
Step 1: Rewrite: sin⁡−1(sin⁡x)=π3−1\sin^{-1}(\sin x)=\dfrac{\pi}{3}-1. Step 2: Case 1: x∈[−π2,π2]x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]. Here sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x)=x:
x=π3−1≈1.047−1=0.047x=\dfrac{\pi}{3}-1\approx 1.047-1=0.047
 Since π3−1∈[−π2,π2]\dfrac{\pi}{3}-1\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], this is valid. Step 3: Case 2: x∈[π2,3π2]x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]. Here sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x)=\pi-x:
π−x=π3−1  ⇒  x=π−π3+1=2π3+1\pi-x=\dfrac{\pi}{3}-1\;\Rightarrow\;x=\pi-\dfrac{\pi}{3}+1=\dfrac{2\pi}{3}+1
 Check: 2π3+1≈2.094+1=3.094\dfrac{2\pi}{3}+1\approx 2.094+1=3.094, and 3π2≈4.71\dfrac{3\pi}{2}\approx 4.71. So x∈[π2,3π2]x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]. Valid. Step 4: Two solutions: x=π3−1x=\dfrac{\pi}{3}-1 and x=2π3+1x=\dfrac{2\pi}{3}+1. Answer: 22
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