a.
\(\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-\dfrac{1}{2}sin2x\right)-\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx-1\right)\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-1\right)\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\sin2x=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(\left(3tan^2x-2\sqrt{3}tanx+1\right)+\left(4sin^2x-4sinx+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{3}tanx-1\right)^2+\left(2sinx-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}tanx=\dfrac{1}{\sqrt{3}}\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
c.
\(\Leftrightarrow4cos4x\left(1+cos4x\right)+\sqrt{1-cos3x}+1=0\)
\(\Leftrightarrow4cos^24x+4cos4x+1+\sqrt{1-cos3x}=0\)
\(\Leftrightarrow\left(2cos4x+1\right)^2+\sqrt{1-cos3x}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2cos4x+1=0\\1-cos3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16cos^4x-16cos^2x+3=0\\4cos^3x-3cosx-1=0\end{matrix}\right.\)
\(\Rightarrow cosx=-\dfrac{1}{2}\)
\(\Rightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)
