2sin5x-3cos5x-2=0 giải giùm kik với

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1. 2sin5x-3cos5x-2=0

   <=>2sin5x-3cos5x=2

   <=>2/{\sqrt{13}} sin5x – 3/{\sqrt{13}} cos5x = 2/{\sqrt{13}}

   Gọi 2/{\sqrt{13}} là cos \alpha , 3/{\sqrt{13}} là sin\alpha ta có:

   <=> cos \alpha sin5x – sin \alpha cos5x= cos \alpha

   <=>sin(5x-\alpha)=sin(\pi/2 – \alpha)  

   <=> \(\left[ \begin{array}{l}5x-\alpha=\dfrac{\pi}{2}-\alpha+k2\pi\\5x-\alpha=\pi-\dfrac{\pi}{2}+\alpha+k2\pi\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}5x=\dfrac{\pi}{2}+k2\pi\\5x=\dfrac{\pi}{2}+2\alpha+k2\pi\end{array} \right.\) 

   <=> <=> \(\left[ \begin{array}{l}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\5x=\dfrac{\pi}{10}+\dfrac{2\alpha}{5}+\dfrac{k2\pi}{5}\end{array} \right.\) (k∈Z)

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