The number of solutions of the equation text{ | vedclass

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(B) Given equation: $	ext{sgn}(\sin x) = \sin^2 x + 2\sin x + 	ext{sgn}(\sin^2 x)$.Since $\sin^2 x > 0$ for $\sin x 
eq 0$,$	ext{sgn}(\sin^2 x) =

Vedclass · Accepted Answer

$6$

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(B) Given equation: $	ext{sgn}(\sin x) = \sin^2 x + 2\sin x + 	ext{sgn}(\sin^2 x)$.Since $\sin^2 x > 0$ for $\sin x 
eq 0$,$	ext{sgn}(\sin^2 x) = 1$ when $\sin x 
eq 0$.Case $1$: If $\sin x > 0$,then $	ext{sgn}(\sin x) = 1$. The equation becomes $1 = \sin^2 x + 2\sin x + 1$,which simplifies to $\sin^2 x + 2\sin x = 0$,or $\sin x(\sin x + 2) = 0$. Since $\sin x > 0$,this yields no solutions.Case $2$: If $\sin x Case $3$: If $\sin x = 0$,then $	ext{sgn}(0) = 0$ and $	ext{sgn}(\sin^2 x) = 0$. The equation becomes $0 = 0 + 0 + 0$,which is $0 = 0$. This is true for all $x = n\pi$ where $n \in \mathbb{Z}$.In the interval $\left[ -\frac{5\pi}{2}, \frac{7\pi}{2} ight]$,the values of $x$ such that $\sin x = 0$ are $x \in \{ -2\pi, -\pi, 0, \pi, 2\pi, 3\pi \}$.Counting these,we have $6$ solutions.

The number of solutions of the equation $\text{sgn}(\sin x) = \sin^2 x + 2\sin x + \text{sgn}(\sin^2 x)$ in the interval $\left[ -\frac{5\pi}{2}, \frac{7\pi}{2} \right]$ is (where $\text{sgn}(\cdot)$ denotes the signum function):

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