The number of solutions of sin 2x + cos 4x = | vedclass

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(C) Given the equation $\sin 2x + \cos 4x = 2$. We know that the range of $\sin 	heta$ is $[-1, 1]$ and the range of $\cos 	heta$ is $[-1, 1]$. For

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$0$

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(C) Given the equation $\sin 2x + \cos 4x = 2$. We know that the range of $\sin 	heta$ is $[-1, 1]$ and the range of $\cos 	heta$ is $[-1, 1]$. For the sum of two functions to be $2$,both functions must simultaneously reach their maximum value of $1$. Thus,we require $\sin 2x = 1$ and $\cos 4x = 1$. From $\sin 2x = 1$,we have $2x = 2n\pi + \frac{\pi}{2}$,which implies $x = n\pi + \frac{\pi}{4}$. For $x \in [-\pi, \pi]$,the possible values are $x = -\frac{3\pi}{4}$ and $x = \frac{\pi}{4}$. Now,check these values in $\cos 4x = 1$: If $x = -\frac{3\pi}{4}$,then $4x = -3\pi$,and $\cos(-3\pi) = -1 
eq 1$. If $x = \frac{\pi}{4}$,then $4x = \pi$,and $\cos(\pi) = -1 
eq 1$. Since no value of $x$ satisfies both equations simultaneously,the number of solutions is $0$.

The number of solutions of $\sin 2x + \cos 4x = 2$ in the interval $[-\pi, \pi]$ is

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