Question

Resolva a equação: sen(x+PI/3) - sen(x-pi/3) = √3/2 , no intervalo [0,2pi[

Answer 1

Olá Luiz.

Temos que utilizar somente a seguinte ferramenta trigonométrica:

$\displaystyle \sin(A+B)=\sin A \cos B + \cos A \sin B \\\\ \sin(A-B)= \sin A \cos B - \cos A \sin B$

Prosseguindo:

$\displaystyle \sin(x+\frac{\pi}{3})-\sin(x-\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \\\\\\ \sin(x)\cos(\frac{\pi}{3})+\cos(x)\sin(\frac{\pi}{3})- \bigg( \sin(x)\cos(\frac{\pi}{3})-\cos(x)\sin(\frac{\pi}{3}) \bigg)=\frac{\sqrt{3}}{2} \\\\\\ \sin(x)\cos(\frac{\pi}{3})+\cos(x)\sin(\frac{\pi}{3})- \sin(x)\cos(\frac{\pi}{3})+\cos(x)\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2} \\\\\\ 2\cos(x)\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Já que o seno de π/3 é igual à √3/2, ficamos com:

$\displaystyle 2\cos x \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \\\\\\ \sqrt{3} \cos x = \frac{\sqrt{3}}{2} \\\\\\ \cos x = \frac{1}{2} \\\\\\ x = \cos^{-1} \bigg(\frac{1}{2} \bigg) \\\\\\ x = \pm \frac{\pi}{3} \, \, \mathsf{rad} \, \, \, \mathsf{ou} \, \, \, \pm \frac{5 \pi}{3} \, \, \mathsf{rad}$