The sum of solutions of the equation frac{cos | vedclass

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Question

(A) Given the equation $\frac{\cos x}{1+\sin x} = |	an 2x|$.Note that $\frac{\cos x}{1+\sin x} = 	an\left(\frac{\pi}{4} - \frac{x}{2}ight)$.So,$

Vedclass · Accepted Answer

$-\frac{11 \pi}{30}$

Vedclass · Answer

(A) Given the equation $\frac{\cos x}{1+\sin x} = |	an 2x|$.Note that $\frac{\cos x}{1+\sin x} = 	an\left(\frac{\pi}{4} - \frac{x}{2}ight)$.So,$	an\left(\frac{\pi}{4} - \frac{x}{2}ight) = |	an 2x|$.Since the $LHS$ must be non-negative,$	an\left(\frac{\pi}{4} - \frac{x}{2}ight) \ge 0$,which implies $\frac{\pi}{4} - \frac{x}{2} \in [0, \frac{\pi}{2})$,so $x \in (-\frac{\pi}{2}, \frac{\pi}{2}]$.Squaring both sides,$	an^2\left(\frac{\pi}{4} - \frac{x}{2}ight) = 	an^2 2x$.This implies $	an 2x = \pm 	an\left(\frac{\pi}{4} - \frac{x}{2}ight)$.Case $1$: $2x = n\pi + \left(\frac{\pi}{4} - \frac{x}{2}ight)$ $\Rightarrow \frac{5x}{2} = n\pi + \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{5} + \frac{\pi}{10}$.For $n=0, x=\frac{\pi}{10}$. For $n=-1, x=-\frac{3\pi}{10}$.Case $2$: $2x = n\pi - \left(\frac{\pi}{4} - \frac{x}{2}ight)$ $\Rightarrow \frac{3x}{2} = n\pi - \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{3} - \frac{\pi}{6}$.For $n=0, x=-\frac{\pi}{6}$.Checking the condition $	an\left(\frac{\pi}{4} - \frac{x}{2}ight) \ge 0$: for $x = \frac{\pi}{10}, -\frac{3\pi}{10}, -\frac{\pi}{6}$,the condition holds.The sum of solutions is $\frac{\pi}{10} - \frac{3\pi}{10} - \frac{\pi}{6} = -\frac{2\pi}{10} - \frac{\pi}{6} = -\frac{\pi}{5} - \frac{\pi}{6} = -\frac{11\pi}{30}$.

The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2 x|$,where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) - \left\{\frac{\pi}{4}, -\frac{\pi}{4}\right\}$,is:

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