Đáp án:
`alpha+beta=(3pi)/5`
Giải thích các bước giải:
`sin3x-cos2x=0`
`<=>sin3x=cos2x`
`<=>sin3x=sin(pi/2-2x)`
`<=>`\(\left[ \begin{array}{l}3x=\dfrac{\pi}{2}-2x+k2\pi\\3x=\pi-(\dfrac{\pi}{2}-2x)+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\5x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{array} \right.\)`(kinZZ)`
`=>alpha=\frac{\pi}{10},beta=pi/2`
`=>alpha+beta=\frac{\pi}{10}+pi/2=(3pi)/5`
