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)  

   <=> $[\begin{array}{l}5x-\alpha ={\displaystyle \frac{\pi }{2}}-\alpha +k2\pi \\ 5x-\alpha =\pi -{\displaystyle \frac{\pi }{2}}+\alpha +k2\pi \end{array}$ 

   <=> $[\begin{array}{l}5x={\displaystyle \frac{\pi }{2}}+k2\pi \\ 5x={\displaystyle \frac{\pi }{2}}+2\alpha +k2\pi \end{array}$ 

   <=> <=> $[\begin{array}{l}x={\displaystyle \frac{\pi }{10}}+{\displaystyle \frac{k2\pi }{5}}\\ 5x={\displaystyle \frac{\pi }{10}}+{\displaystyle \frac{2\alpha }{5}}+{\displaystyle \frac{k2\pi }{5}}\end{array}$ (k∈Z)

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