The number of all the values of x for which t | vedclass

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(C) Given $f(x) = \sin x + \frac{1-	an^2 x}{1+	an^2 x}$.Using the identity $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we have $f(x) = \sin x + \cos 2

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$2$

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(C) Given $f(x) = \sin x + \frac{1-	an^2 x}{1+	an^2 x}$.Using the identity $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we have $f(x) = \sin x + \cos 2x$.Since $\cos 2x = 1 - 2\sin^2 x$,we can write $f(x) = \sin x + 1 - 2\sin^2 x$.Let $t = \sin x$,where $t \in [-1, 1]$. Then $g(t) = -2t^2 + t + 1$.To find the maximum,we find the derivative $g'(t) = -4t + 1$.Setting $g'(t) = 0$,we get $t = \frac{1}{4}$.Since $g''(t) = -4 We need to find the number of values of $x \in [0, 2\pi]$ such that $\sin x = \frac{1}{4}$.Since $\frac{1}{4} > 0$,$\sin x = \frac{1}{4}$ has two solutions in the interval $[0, 2\pi]$ (one in the first quadrant and one in the second quadrant).Thus,there are $2$ such values of $x$.

The number of all the values of $x$ for which the function $f(x)=\sin x+\frac{1-\tan ^2 x}{1+\tan ^2 x}$ attains its maximum value on $[0, 2\pi]$ is

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