Giải pt dạng asinx +- bcosx = c Sin3x-cos3x =1

Sin3x-cos3x =1

sin( 3x - pi/3)= √2 /2

3x- pi/4= pi/4 + k2pi hoặc 3x- pi/4 = 3pi/4 + k2pi

x= pi/6 + k2pi/3 hoặc x= pi/3 + k2pi/3

$sin3x-cos3x=-1$

$⇔\sqrt[]{2}sin(3x-\dfrac{\pi}{4})=-1$

$⇔sin(3x-\dfrac{\pi}{4})=\dfrac{-1}{\sqrt[]{2}}=sin(-\dfrac{\pi}{4})$

$⇔$\(\left[ \begin{array}{l}3x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\3x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}3x=k2\pi\\3x=\dfrac{3\pi}{2}+k2\pi\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{2}+\dfrac{k2\pi}{3}\end{array} \right.\)