The number of real roots of the equation sin^ | vedclass

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(B) Given equation: $\sin^{2020} x - \cos^{2020} x + 2019 = 2020$ $\Rightarrow \sin^{2020} x = 1 + \cos^{2020} x$ Since the range of $\sin^{2020} x$ i

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$3$

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(B) Given equation: $\sin^{2020} x - \cos^{2020} x + 2019 = 2020$ $\Rightarrow \sin^{2020} x = 1 + \cos^{2020} x$ Since the range of $\sin^{2020} x$ is $[0, 1]$ and the range of $1 + \cos^{2020} x$ is $[1, 2]$,the equality holds only when $\sin^{2020} x = 1$ and $\cos^{2020} x = 0$. This implies $\sin x = \pm 1$ and $\cos x = 0$. In the interval $\left(-\frac{3\pi}{2}, \frac{5\pi}{2}\right)$,the values of $x$ satisfying $\cos x = 0$ are $x = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}$. Checking these values in $\sin^{2020} x = 1$: For $x = -\frac{\pi}{2}$,$\sin(-\frac{\pi}{2}) = -1$,so $(-1)^{2020} = 1$ (True). For $x = \frac{\pi}{2}$,$\sin(\frac{\pi}{2}) = 1$,so $(1)^{2020} = 1$ (True). For $x = \frac{3\pi}{2}$,$\sin(\frac{3\pi}{2}) = -1$,so $(-1)^{2020} = 1$ (True). Thus,there are $3$ real roots.

The number of real roots of the equation $\sin^{2020} x - \cos^{2020} x + 2019 = 2020$ in the interval $\left(-\frac{3\pi}{2}, \frac{5\pi}{2}\right)$ is:

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