ĐK: \(x\ne\dfrac{\pi}{2}+k\pi;x\ne\pi+k2\pi\)
\(\dfrac{\sqrt{3}sin2x\left(1+2cosx\right)+cos3x}{2cosx+2cos^2x}=1\)
\(\Leftrightarrow\dfrac{2\sqrt{3}sinx.cosx\left(1+2cosx\right)+4cos^3x-3cosx}{2cosx+2cos^2x}=1\)
\(\Leftrightarrow\dfrac{2\sqrt{3}sinx\left(1+2cosx\right)+4cos^2x-3}{2+2cosx}=1\)
\(\Leftrightarrow2\sqrt{3}sinx\left(1+2cosx\right)+4cos^2x-3=2+2cosx\)
\(\Leftrightarrow2\sqrt{3}sinx+4\sqrt{3}sinx.cosx+4cos^2x-2cosx-5=0\)
\(\Leftrightarrow2\sqrt{3}sinx-2cosx+2\sqrt{3}sin2x+2cos2x-3=0\)
\(\Leftrightarrow2\left(\sqrt{3}sinx-cosx\right)+2\left(\sqrt{3}sin2x+cos2x\right)-3=0\)
\(\Leftrightarrow4sin\left(x-\dfrac{\pi}{6}\right)+4cos\left(2x-\dfrac{\pi}{3}\right)-3=0\)
\(\Leftrightarrow4sin\left(x-\dfrac{\pi}{6}\right)+4-8sin^2\left(x-\dfrac{\pi}{6}\right)-3=0\)
\(\Leftrightarrow8sin^2\left(x-\dfrac{\pi}{6}\right)-4sin\left(x-\dfrac{\pi}{6}\right)-1=0\)
Đến đây thì dễ rồi.
