`sin(x+1/3)+cos(x-1)=0`
`<=>sin(x+1/3)=-cos(x-1)`
`<=>sin(x+1/3)=sin(x-\pi/2-1)`
`<=>[(x+1/3=x-\pi/2-1+k2\pi),(x+1/3=[3\pi]/2+1-x+k2\pi):}`
`<=>[(0x=-\pi/2-4/3+k2\pi (VN)),(x=1/3+[3\pi]/4+k\pi):}`
`<=>x=1/3+[3\pi]/4+k\pi` `(k in ZZ)`
\(\Leftrightarrow sin\left(x+\dfrac{1}{3}\right)=-cos\left(x-1\right)=-sin\left(\dfrac{pi}{2}-x+1\right)\)
\(\Leftrightarrow sin\left(x+\dfrac{1}{3}\right)=sin\left(-\dfrac{pi}{2}+x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=-\dfrac{pi}{2}+x-1+k2pi\\x+\dfrac{1}{3}=\dfrac{3}{2}pi-x+1+k2pi\end{matrix}\right.\Leftrightarrow2x=\dfrac{3}{2}pi+\dfrac{2}{3}+k2pi\)
hay \(x=\dfrac{3}{4}pi+\dfrac{1}{3}+kpi\)
