1.
Đặt \(\frac{x}{3}=t\) pt trở thành:
\(cos4t=sin^23t\Leftrightarrow2cos4t=1-cos6t\)
\(\Leftrightarrow cos6t+2cos4t-1=0\)
\(\Leftrightarrow4cos^32t-3cos2t+2\left(2cos^22t-1\right)-1=0\)
\(\Leftrightarrow4cos^32t+2cos^22t-3cos2t-3=0\)
\(\Leftrightarrow\left(cos2t-1\right)\left(4cos^22t+6cos2t+3\right)=0\)
\(\Leftrightarrow cos2t=1\Leftrightarrow cos\frac{2x}{3}=1\)
\(\Leftrightarrow\frac{2x}{3}=k2\pi\Leftrightarrow x=k3\pi\)
2.
\(\Leftrightarrow4cos^3x-3cosx-\left(1-2sin^2x\right)+9sinx-4=0\)
\(\Leftrightarrow cosx\left(4cos^2x-3\right)+2sin^2x+9sinx-5=0\)
\(\Leftrightarrow cosx\left(4\left(1-sin^2x\right)-3\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow cosx\left(1-4sin^2x\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow\left(cosx+2sinx.cosx\right)\left(1-2sinx\right)-\left(1-2sinx\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow\left(1-2sinx\right)\left(cosx-sinx+2sinx.cosx-5\right)=0\)
\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)+sin2x-5\right)=0\)
\(\Leftrightarrow1-2sinx=0\) (do \(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\le\sqrt{2};sin2x\le1\) nên ngoặc sau luôn âm)
\(\Leftrightarrow sinx=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
