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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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:

Similar Questions

What is the number of values of $x$ in the interval $[0, 5\pi]$ that satisfy the equation $3\sin^2x - 7\sin x + 2 = 0$?

The number of values of $x$ for which $\sin 2x + \cos 4x = 2$ is

The equation $\sqrt{3} \sin x + \cos x = 4$ has

If $1 + \cot \theta = \text{cosec} \theta$,then the general value of $\theta$ is

If $\sqrt{3} \tan 2\theta + \sqrt{3} \tan 3\theta + \tan 2\theta \tan 3\theta = 1$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module