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SIMAK UI 2010

Solusi SIMAK UI 2010 Kode 509 [IPA] – Nomor 8


Jika\ \frac{\pi}{2}\le x\le \frac{3\pi}{2}\ memenuhi\ persamaan
\frac{1-sinx}{cosx}+\frac{2cosx}{1+sinx}=8cosx
maka\ nilai\ sinx=....
(A)\quad -1
(B)\quad -\frac{5}{8}
(C)\quad 0
(D)\quad \frac{5}{8}
(E)\quad 1
Jawaban: B

\frac{1-sinx}{cosx}+\frac{2cosx}{1+sinx}=8cosx
\frac{1-sin^{2}x+2cos^{2}x}{cosx(1+sinx)}=8cosx
1-sin^{2}x+2cos^{2}x=8cos^{2}x(1+sinx)
cos^{2}x+2cos^{2}x=8cos^{2}x+8cos^{2}xsinx
3cos^{2}x=8cos^{2}x+8cos^{2}xsinx
-5cos^{2}x=8cos^{2}xsinx
-5=8sinx
sinx=-\frac{5}{8}

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