SBMPTN Zone : Sistem Persamaan Trigonometri

Berikut ini adalah kumpulan soal mengenai Sistem Persamaan Trigonometri Tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Jika x dan y memenuhi persamaan
\begin{cases} x-y=\dfrac{\pi}6\\[8pt] \cos x-\sqrt3\cos y=-\dfrac12 \end{cases}
dengan {x\in[-\pi,\pi]}, {y\in[-\pi,\pi]}, maka nilai {\cos(x+y)} yang mungkin adalah

1. -\dfrac12\sqrt2
2. \dfrac12\sqrt2
3. -\dfrac12\sqrt3

4. \dfrac13\sqrt5
5. \dfrac13\sqrt3

SUMBER

\(\begin{aligned} x-y&=\dfrac{\pi}6\\ x&=\dfrac{\pi}6+y \end{aligned}\)

\(\begin{aligned} \cos x-\sqrt3\cos y&=-\dfrac12\\ \cos\left(\dfrac{\pi}6+y\right)-\sqrt3\cos y&=-\dfrac12\\ \cos\dfrac{\pi}6\cos y-\sin\dfrac{\pi}6\sin y-\sqrt3\cos y&=-\dfrac12\\ \dfrac12\sqrt3\cos y-\dfrac12\sin y-\sqrt3\cos y&=-\dfrac12\\ -\dfrac12\sqrt3\cos y-\dfrac12\sin y&=-\dfrac12\\ \dfrac12\sqrt3\cos y+\dfrac12\sin y&=\dfrac12\\ \sin\dfrac{\pi}3\cos y+\cos\dfrac{\pi}3\sin y&=\dfrac12\\ \sin\left(\dfrac{\pi}3+y\right)&=\sin\dfrac{\pi}6 \end{aligned}\)

• \dfrac{\pi}3+y=\dfrac{\pi}6
  \(\begin{aligned} y&=\dfrac{\pi}6-\dfrac{\pi}3\\ &=-\dfrac{\pi}6\end{aligned}\)
  
  \(\begin{aligned} x&=\dfrac{\pi}6+y\\ x&=\dfrac{\pi}6+\left(-\dfrac{\pi}6\right)\\ &=0 \end{aligned}\)
  
  \(\begin{aligned} \cos(x+y)&=\cos\left(0-\dfrac{\pi}6\right)\\ &=\cos\left(-\dfrac{\pi}6\right)\\ &=\cos\dfrac{\pi}6\\ &=\dfrac12\sqrt3 \end{aligned}\)
• \dfrac{\pi}3+y=\dfrac{5\pi}6
  \(\begin{aligned} y&=\dfrac{5\pi}6-\dfrac{\pi}3\\ &=\dfrac{3\pi}6\end{aligned}\)
  
  \(\begin{aligned} x&=\dfrac{\pi}6+y\\ x&=\dfrac{\pi}6+\dfrac{3\pi}6\\ &=\dfrac{4\pi}6 \end{aligned}\)
  
  \(\begin{aligned} \cos(x+y)&=\cos\left(\dfrac{4\pi}6+\dfrac{3\pi}6\right)\\ &=\cos\dfrac{7\pi}6\\ &=\cos\dfrac{\pi}6\\ &=\boxed{\boxed{-\dfrac12\sqrt3}} \end{aligned}\)

Jadi, nilai {\cos(x+y)} yang mungkin adalah -\dfrac12\sqrt3.
JAWAB: C