Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Find Number Solutions Equation

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

Question
Find the number of solutions of the equation sin⁡−1(4sin⁡2θ+sin⁡θ)+cos⁡−1(−1+6sin⁡θ)=π2\sin^{-1}(4\sin^{2}\theta+\sin\theta)+\cos^{-1}(-1+6\sin\theta)=\dfrac{\pi}{2} in θ∈[0,5π]\theta\in[0,5\pi].
Solution
Answer: 9
Step 1: Using the identity sin⁡−1A+cos⁡−1B=π2  ⟺  A=B\sin^{-1}A+\cos^{-1}B=\dfrac{\pi}{2}\iff A=B (when both are in valid ranges):
4sin⁡2θ+sin⁡θ=−1+6sin⁡θ4\sin^{2}\theta+\sin\theta=-1+6\sin\theta
 Step 2: Rearrange:
4sin⁡2θ−5sin⁡θ+1=04\sin^{2}\theta-5\sin\theta+1=0
 Step 3: Factor:
(4sin⁡θ−1)(sin⁡θ−1)=0  ⇒  sin⁡θ=14 or sin⁡θ=1(4\sin\theta-1)(\sin\theta-1)=0\;\Rightarrow\;\sin\theta=\dfrac{1}{4}\text{ or }\sin\theta=1
 Step 4: Count solutions in [0,5π][0,5\pi]: For sin⁡θ=1\sin\theta=1: θ=π2+2kπ\theta=\dfrac{\pi}{2}+2k\pi. In [0,5π][0,5\pi]: π2,5π2,9π2\dfrac{\pi}{2},\dfrac{5\pi}{2},\dfrac{9\pi}{2}. Total: 33 solutions. For sin⁡θ=14\sin\theta=\dfrac{1}{4}: Two solutions per period 2π2\pi. In [0,5π][0,5\pi] (length 5π5\pi, so 2.52.5 periods): 55 or 66 solutions. Counting carefully: [0,2π][0,2\pi] has 22, [2π,4π][2\pi,4\pi] has 22, [4π,5π][4\pi,5\pi] has 22 (since sin⁡\sin is positive on [4π,5π][4\pi,5\pi]). Total: 66 solutions. Step 5: Total: 3+6=93+6=9. Answer: 99
Solution working
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