\(sinx+\left(\sqrt{3}-2\right)cosx=1\)
\(\Leftrightarrow sinx+\sqrt{3}cosx=2cosx+1\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=cosx+\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=2cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow2sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=2cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)-cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\right].cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow\left[sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)-sin\left(\dfrac{2\pi}{3}-\dfrac{x}{2}\right)\right].cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow cos\dfrac{5\pi}{12}.sin\left(x-\dfrac{\pi}{2}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)
...
