\(\Leftrightarrow4\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)^2=5+cos\left(4x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow4sin^2\left(x+\dfrac{\pi}{3}\right)=5+cos\left(4x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow2-2cos\left(2x+\dfrac{2\pi}{3}\right)=5+cos\left(4x+\dfrac{\pi}{3}\right)\)
Đặt \(2x+\dfrac{2\pi}{3}=t\Rightarrow2x=t-\dfrac{2\pi}{3}\Rightarrow4x+\dfrac{\pi}{3}=2\left(t-\dfrac{2\pi}{3}\right)+\dfrac{\pi}{3}=2t-\pi\)
Pt trở thành:
\(-2cost=3+cos\left(2t-\pi\right)=3-cos2t\)
\(\Leftrightarrow cos2t-2cost-3=0\)
\(\Leftrightarrow cos^2t-cost-2=0\Rightarrow\left[{}\begin{matrix}cost=-1\\cost=2\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pi+k2\pi\Rightarrow2x+\dfrac{2\pi}{3}=\pi+k2\pi\)
\(\Rightarrow x=\dfrac{\pi}{6}+k\pi\)
\(\Rightarrow\) Pt có 4 nghiệm dương bé hơn 10 là \(\dfrac{\pi}{6};\dfrac{7\pi}{6};\dfrac{13\pi}{6};\dfrac{19\pi}{6}\)
