a, \(sin^2x+sinx.cosx=\dfrac{1}{2}\)
\(\Leftrightarrow2sin^2x-1+2sinx.cosx=0\)
\(\Leftrightarrow sin2x-cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow2x-\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)
b, ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(4sinx+6cosx=\dfrac{1}{cosx}\)
\(\Leftrightarrow\dfrac{4sinx}{cosx}+6=\dfrac{1}{cos^2x}\)
\(\Leftrightarrow4tanx+6=tan^2x+1\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tanx-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan5+k\pi\end{matrix}\right.\)
c, \(2sin^25x+sin10x+4cos^25x=3\)
\(\Leftrightarrow2sin^25x-1+sin10x+4cos^25x-2=0\)
\(\Leftrightarrow sin10x+cos10x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(10x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow10x+\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{40}+\dfrac{k\pi}{10}\)
d, \(\left(\sqrt{3}+1\right)\left(sinx+cosx\right)cosx=1\)
\(\Leftrightarrow sinx.cosx+cos^2x=\dfrac{1}{\sqrt{3}+1}\)
\(\Leftrightarrow2sinx.cosx+2cos^2x=\dfrac{2}{\sqrt{3}+1}\)
\(\Leftrightarrow sin2x+cos2x=\sqrt{3}-2\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{3}-2\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}-2}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=arcsin\left(\dfrac{\sqrt{3}-2}{\sqrt{2}}\right)+k2\pi\\2x+\dfrac{\pi}{4}=\pi-arcsin\left(\dfrac{\sqrt{3}-2}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+\dfrac{1}{2}arcsin\left(\dfrac{\sqrt{3}-2}{\sqrt{2}}\right)+k\pi\\x=\dfrac{3\pi}{8}-\dfrac{1}{2}arcsin\left(\dfrac{\sqrt{3}-2}{\sqrt{2}}\right)+k\pi\end{matrix}\right.\)
