The number of solutions of the equation $1 + \sin x \cdot \sin^2 \frac{x}{2} = 0$ in $[-\pi, \pi]$ is

Question

Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>, \theta \in(0,2 \pi)$, holds If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

Answer 1

To solve the given trigonometric equation $1 + \sin x \cdot \sin^2 \frac{x}{2} = 0$ in the interval $[-\pi, \pi]$, we need to find the values of $x$ that satisfy the equation. Let's break down the problem step-by-step:

The equation is given as: $1 + \sin x \cdot \sin^2 \frac{x}{2} = 0$
Rearrange the equation to isolate the trigonometric expression: $\sin x \cdot \sin^2 \frac{x}{2} = -1$
Notice that: $|\sin x| \leq 1$ and $|\sin^2 \frac{x}{2}| \leq 1$
Therefore, $|\sin x \cdot \sin^2 \frac{x}{2}| \leq 1$. Hence, $\sin x \cdot \sin^2 \frac{x}{2}$ cannot be equal to $-1$ based on the range of sine functions.

Since it is impossible for the product of two terms, each bounded within $[-1, 1]$, to equal $-1$, the equation has no solutions.

Conclusion: The number of solutions of the equation in the interval $[-\pi, \pi]$ is zero.

The number of solutions of the equation \(1 + \sin x \cdot \sin^2 \frac{x}{2} = 0\) in \([-\pi, \pi]\) is

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The Correct Option is A

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