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SIMAK UI 2010

Solusi SIMAK UI 2010 Kode 209 Nomor 3


Nilai\ terbesar\ x\ dalam\ 0<x<2\pi\ sehingga\ cos \big(2x-\frac{\pi}{2}\big)=\sqrt{3}\ sin \big(2x-\frac{\pi}{2}\big)\\adalah\ ....\\\\(A) \,\,\, \frac{13\pi}{12}\\\\(B) \,\,\, \frac{7\pi}{6}\\\\(C) \,\,\, \frac{8\pi}{6}\\\\(D) \,\,\, \frac{19\pi}{12}\\\\(E) \,\,\, \frac{11\pi}{6}
Jawaban: E

\frac{sin \big(2x-\frac{\pi}{2}\big)}{cos \big(2x-\frac{\pi}{2})}=\frac{1}{\sqrt{3}}\\\\tan \big(2x-\frac{\pi}{2}\big)=tan\ \frac{\pi}{6}\\\\2x-\frac{\pi}{2}=\frac{\pi}{6}+k\pi\\\\2x=\frac{\pi}{2}+\frac{\pi}{6}+k\pi\\\\2x=\frac{4\pi}{6}+k\pi\\\\2x=\frac{2\pi}{3}+k\pi\\\\x=\frac{\pi}{3}+k.\frac{\pi}{2}\\\\Dalam\ interval\ 0<x<2\pi\ dipilih:\\\\k=0 \quad \rightarrow \quad x=\frac{\pi}{3}\\\\k=1 \quad \rightarrow \quad x=\frac{\pi}{3}+\frac{\pi}{2}=\frac{5\pi}{6}\\\\k=2 \quad \rightarrow \quad x=\frac{\pi}{3}+\pi=\frac{4\pi}{3}=\frac{8\pi}{6}\\\\k=3 \quad \rightarrow \quad x=\frac{\pi}{3}+\pi=\frac{3\pi}{2}=\bf{\frac{11\pi}{6}}

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