The number of solutions of the equation 2x + | vedclass

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(B) The given equation is $2x + 3 	an x = \pi$. This can be rewritten as $	an x = \frac{\pi - 2x}{3} = \frac{\pi}{3} - \frac{2x}{3}$. To find the nu

Vedclass · Accepted Answer

$5$

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(B) The given equation is $2x + 3 	an x = \pi$. This can be rewritten as $	an x = \frac{\pi - 2x}{3} = \frac{\pi}{3} - \frac{2x}{3}$. To find the number of solutions,we plot the graphs of $y = 	an x$ and $y = -\frac{2}{3}x + \frac{\pi}{3}$ on the interval $x \in [-2\pi, 2\pi]$. The function $y = 	an x$ has vertical asymptotes at $x = \pm \frac{\pi}{2}$ and $x = \pm \frac{3\pi}{2}$. The line $y = -\frac{2}{3}x + \frac{\pi}{3}$ has a negative slope and passes through $(0, \frac{\pi}{3})$. By observing the intersection points of the graph of $	an x$ and the straight line,we can see that there are $5$ points of intersection within the given domain. Therefore,the number of solutions is $5$.

The number of solutions of the equation $2x + 3 \tan x = \pi$,where $x \in [-2\pi, 2\pi] - \left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$,is

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