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[tex3]x[/tex3]
[tex3]0 < x \leq \frac{\pi }{2}[/tex3]
[tex3]\text{sen}(4x)\cdot\cot(x)=\frac{3+2\sqrt{3}}{2}[/tex3]
[tex3]x=\frac{\pi }{12}[/tex3]
[tex3]x=\frac{\pi }{6}[/tex3]
[tex3]\sin(4x)\cdot\cot(x)=\frac{3+2\sqrt{3}}{2}[/tex3]
[tex3]\sin(4x)[/tex3]
[tex3]\sin(4x)=[/tex3]
[tex3]=\sin(2x+2x)=[/tex3]
[tex3]=2\sin 2x\cos 2x=[/tex3]
[tex3]=2\sin(x+x)\cos(x+x)=[/tex3]
[tex3]=2(2\sin x\cos x)(\cos^2x-\sin^2x)=[/tex3]
[tex3]=2(2\sin x\cos x)(\cos^2x-(1-\cos^2x))=[/tex3]
[tex3]=2(2\sin x\cos x)(2\cos^2x-1)[/tex3]
[tex3]\sin(4x)\cdot\cot(x)=[/tex3]
[tex3]2(2\sin x\cos x)(2\cos^2x-1)\cdot\cot(x)=[/tex3]
[tex3]=2(2\cancel{\sin x}\cos x)(2\cos^2x-1)\cdot\frac{\cos x}{\cancel{\sin x}}=[/tex3]
[tex3]=2(2\cos^2x)(2\cos^2x-1)=[/tex3]
[tex3]=8\cos^4x-4\cos^2x[/tex3]
[tex3]\sin(4x)\cdot\cot(x)[/tex3]
[tex3]8\cos^4x-4\cos^2x=\frac{3+2\sqrt{3}}{2}[/tex3]
[tex3]16\cos^4x-8\cos^2x=3+2\sqrt{3}[/tex3]
[tex3]16\cos^4x-8\cos^2x-(3+2\sqrt{3})=0[/tex3]
[tex3]\cos^2x=y[/tex3]
[tex3]16y^2-8y-(3+2\sqrt{3})=0\rightarrow\begin{cases}y_1=\frac{1+\sqrt{2(2+\sqrt{3}})}{2}\\y_2=\frac{1-\sqrt{2(2+\sqrt{3}})}{2}\end{cases}[/tex3]
[tex3]y=\cos^2x[/tex3]
[tex3]\cos x=\sqrt{\frac{1\pm\sqrt{2(2+\sqrt{3})}}{4}}[/tex3]
[tex3]\cos x=\sqrt{\frac{1+\sqrt{2(2+\sqrt{3})}}{4}}\ (I)[/tex3]
[tex3]\cos\theta=\sqrt{\frac{1+\cos2\theta}{2}}[/tex3]
[tex3]\cos x=\sqrt{\frac{1-\sqrt{2(2+\sqrt{3})}}{4}}\ (II)[/tex3]
[tex3]\cos\theta=\sqrt{\frac{1-\sin2\theta}{2}}[/tex3]
[tex3]\sqrt{2(2+\sqrt{3})}[/tex3]
[tex3]\sqrt{2(2+\sqrt{3})}=\sqrt{4+2\sqrt{3}}=\sqrt{1+2\sqrt{3}+3}=\sqrt{(1+\sqrt{3})^2}=1+\sqrt{3}[/tex3]
[tex3](I)[/tex3]
[tex3]\cos x=\sqrt{\frac{1+(1+\sqrt{3})}{4}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}[/tex3]
[tex3]\cos2x=\frac{\sqrt{3}}{2}\rightarrow 2x=\frac{\pi}{6}\rightarrow x=\frac{\pi}{12}[/tex3]
[tex3](II)[/tex3]
[tex3]\cos x=\sqrt{\frac{1-(1+\sqrt{3})}{4}}=\sqrt{-\frac{\sqrt{3}}{4}}\notin\mathbb{R}[/tex3]
[tex3]S=\left\{\frac{\pi}{12}\right\}[/tex3]
[tex3]\frac{\pi}{6}[/tex3]