ta có \(\sin1\cos x=\dfrac{sin2x}{2};cos2x=\cos^2x-sin^2\)
\(\rightarrow\left(\sqrt{3}+1\right)\left(\sin1\cos x+\cos^2x\right)=1\)
\(\Leftrightarrow\sin x.\cos x+\cos^2x=\dfrac{1}{\sqrt{3}+1}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{1}{2}\cos2x+\dfrac{1}{2}=\dfrac{1}{\sqrt{3}+1}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\sin2x+\cos x\right)=\dfrac{\sqrt{3}-2}{2}\)
\(\Leftrightarrow\sin2x+\cos2x=\sqrt{3}-2\)
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}\sin2x+\dfrac{\sqrt{2}}{2}\cos2x=\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-2\right)\)
\(\Rightarrow\sin\left(\dfrac{n}{4}+2x\right)=\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-2\right)\)
\(\Rightarrow x=\dfrac{1}{2}\left(n-arcsin\left(\dfrac{\sqrt{2}}{2}(\sqrt{3}-2\right)\right)+k2n\overline{\overline{4}}^{12}\)
hay \(x=\dfrac{1}{2}(-\dfrac{n}{4}+n-arcsin(\dfrac{\sqrt{2}}{2}(\sqrt{5}-2))k2n\)
