The sum of solutions of the equation x1+ x = |2x|, x(-2,2)4 is:

Question

The sum of solutions of the equation $\dfrac{\cos x}{1+\sin x} = |	an2x|$, $x\in\left(-\dfrac\pi2,\dfrac\pi2ight)\setminus\left\{\pm\dfrac\pi4ight\}$ is:

Accepted Answer

Writing $\dfrac{\cos x}{1+\sin x} = \dfrac{\cos\tfrac x2-\sin\tfrac x2}{\cos\tfrac x2+\sin\tfrac x2} = \dfrac{1-\tan\tfrac x2}{1+\tan\tfrac x2} = \tan\left(\dfrac\pi4-\dfrac x2\right)$, the equation becomes:

$$\tan^2\left(\tfrac\pi4-\tfrac x2\right) = \tan^22x \implies 2x = n\pi\pm\left(\tfrac\pi4-\tfrac x2\right)$$

This splits into $\dfrac{5x}2 = \left(n+\tfrac14\right)\pi$ or $\dfrac{3x}2 = \left(n-\tfrac14\right)\pi$. Restricting $x$ to $\left(-\tfrac\pi2,\tfrac\pi2\right)$ gives valid integer $n$: $n=-1,0$ (first family) and $n=0$ (second family), yielding $x = -\tfrac{3\pi}{10}, \tfrac\pi{10}, -\tfrac\pi6$.

$$\text{Sum} = -\frac{3\pi}{10}+\frac\pi{10}-\frac\pi6 = -\frac{2\pi}{10}-\frac\pi6 = -\frac\pi5-\frac\pi6 = -\frac{11\pi}{30}$$

Solutions · 1