Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question

Find the number of solutions of the equation

\sin^{-1}(4\sin^{2}\theta+\sin\theta)+\cos^{-1}(-1+6\sin\theta)=\dfrac{\pi}{2}

\theta\in[0,5\pi]

Solution

Answer: 9

Step 1: Using the identity

\sin^{-1}A+\cos^{-1}B=\dfrac{\pi}{2}\iff A=B

(when both are in valid ranges):

4\sin^{2}\theta+\sin\theta=-1+6\sin\theta

Step 2: Rearrange:

4\sin^{2}\theta-5\sin\theta+1=0

Step 3: Factor:

(4\sin\theta-1)(\sin\theta-1)=0\;\Rightarrow\;\sin\theta=\dfrac{1}{4}\text{ or }\sin\theta=1

Step 4: Count solutions in

[0,5\pi]

: For

\sin\theta=1

\theta=\dfrac{\pi}{2}+2k\pi

. In

[0,5\pi]

\dfrac{\pi}{2},\dfrac{5\pi}{2},\dfrac{9\pi}{2}

. Total:

3

solutions. For

\sin\theta=\dfrac{1}{4}

: Two solutions per period

2\pi

. In

[0,5\pi]

(length

5\pi

, so

2.5

periods):

5

6

solutions. Counting carefully:

[0,2\pi]

has

2

[2\pi,4\pi]

has

2

[4\pi,5\pi]

has

2

(since

\sin

is positive on

[4\pi,5\pi]

). Total:

6

solutions. Step 5: Total:

3+6=9

. Answer:

9

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