Toán 11 Sin(3x+pi/3)-cos(5pi/6+3x)=2

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THPT Marie Curie
$\sin \left(3x + \dfrac{\pi}{3}\right) - \cos \left(\dfrac{5\pi}{6} + 3x\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) - \cos \left(\dfrac{\pi}{2} - \left(-\dfrac{\pi}{3} - 3x\right)\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) - \sin \left(-\dfrac{\pi}{3} - 3x\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) + \sin \left(\dfrac{\pi}{3} + 3x\right) = 2$
$\implies 2\sin \left(3x + \dfrac{\pi}{3}\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) = 1$
$\implies 3x + \dfrac{\pi}{3} = \dfrac{\pi}{2} + k2\pi$
$\implies x = \dfrac{\pi}{18} + \dfrac{k2\pi}{3} \,\, (k \in Z)$
 
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