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Inverse Trigonometric Functions: Let Defined Sgn Find Number Solutions Lying

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

Question
Let g:R→Rg:\mathbb{R}\to\mathbb{R} be defined as g(x)=sgn(x2−5x+6)g(x)=\text{sgn}(x^{2}-5x+6). Find the number of solutions of sin⁡x=cos⁡−1(g(sin⁡−1x))\sin x=\cos^{-1}(g(\sin^{-1}x)) lying in [0,314][0,314].
Solution
Answer: 1
Step 1: g(sin⁡−1x)=sgn((sin⁡−1x)2−5sin⁡−1x+6)=sgn((sin⁡−1x−2)(sin⁡−1x−3))g(\sin^{-1}x)=\text{sgn}((\sin^{-1}x)^{2}-5\sin^{-1}x+6)=\text{sgn}((\sin^{-1}x-2)(\sin^{-1}x-3)). Step 2: Domain of sin⁡−1x\sin^{-1}x is [−1,1][-1,1], with range [−π2,π2]≈[−1.57,1.57]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\approx[-1.57,1.57]. Step 3: For sin⁡−1x∈[−π2,π2]\sin^{-1}x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]: both sin⁡−1x−2<0\sin^{-1}x-2<0 and sin⁡−1x−3<0\sin^{-1}x-3<0, so their product is positive. Hence g(sin⁡−1x)=1g(\sin^{-1}x)=1 always (for x∈[−1,1]x\in[-1,1]). Step 4: The equation becomes:
sin⁡x=cos⁡−1(1)=0  ⇒  sin⁡x=0  ⇒  x=nπ,  n∈Z\sin x=\cos^{-1}(1)=0\;\Rightarrow\;\sin x=0\;\Rightarrow\;x=n\pi,\;n\in\mathbb{Z}
 Step 5: But the equation is only meaningful for x∈[−1,1]x\in[-1,1] (the domain of sin⁡−1\sin^{-1}). The only multiple of π\pi in [−1,1][-1,1] is x=0x=0. Step 6: Only 11 solution. Answer: 11
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