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
$$\{\begin{array}{l}x-y={\displaystyle \frac{\pi }{6}}\\ \cos x-\sqrt{3}\cos y=-{\displaystyle \frac{1}{2}}\end{array}$$
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

$\begin{array}{rl}x-y& ={\displaystyle \frac{\pi }{6}}\\ x& ={\displaystyle \frac{\pi }{6}}+y\end{array}$

$\begin{array}{rl}\cos x-\sqrt{3}\cos y& =-{\displaystyle \frac{1}{2}}\\ \cos ({\displaystyle \frac{\pi }{6}}+y)-\sqrt{3}\cos y& =-{\displaystyle \frac{1}{2}}\\ \cos {\displaystyle \frac{\pi }{6}}\cos y-\sin {\displaystyle \frac{\pi }{6}}\sin y-\sqrt{3}\cos y& =-{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}\sqrt{3}\cos y-{\displaystyle \frac{1}{2}}\sin y-\sqrt{3}\cos y& =-{\displaystyle \frac{1}{2}}\\ -{\displaystyle \frac{1}{2}}\sqrt{3}\cos y-{\displaystyle \frac{1}{2}}\sin y& =-{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}\sqrt{3}\cos y+{\displaystyle \frac{1}{2}}\sin y& ={\displaystyle \frac{1}{2}}\\ \sin {\displaystyle \frac{\pi }{3}}\cos y+\cos {\displaystyle \frac{\pi }{3}}\sin y& ={\displaystyle \frac{1}{2}}\\ \sin ({\displaystyle \frac{\pi }{3}}+y)& =\sin {\displaystyle \frac{\pi }{6}}\end{array}$

• \dfrac{\pi}3+y=\dfrac{\pi}6
  $\begin{array}{rl}y& ={\displaystyle \frac{\pi }{6}}-{\displaystyle \frac{\pi }{3}}\\ & =-{\displaystyle \frac{\pi }{6}}\end{array}$
  
  $\begin{array}{rl}x& ={\displaystyle \frac{\pi }{6}}+y\\ x& ={\displaystyle \frac{\pi }{6}}+(-{\displaystyle \frac{\pi }{6}})\\ & =0\end{array}$
  
  $\begin{array}{rl}\cos (x+y)& =\cos (0-{\displaystyle \frac{\pi }{6}})\\ & =\cos (-{\displaystyle \frac{\pi }{6}})\\ & =\cos {\displaystyle \frac{\pi }{6}}\\ & ={\displaystyle \frac{1}{2}}\sqrt{3}\end{array}$
• \dfrac{\pi}3+y=\dfrac{5\pi}6
  $\begin{array}{rl}y& ={\displaystyle \frac{5\pi }{6}}-{\displaystyle \frac{\pi }{3}}\\ & ={\displaystyle \frac{3\pi }{6}}\end{array}$
  
  $\begin{array}{rl}x& ={\displaystyle \frac{\pi }{6}}+y\\ x& ={\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{3\pi }{6}}\\ & ={\displaystyle \frac{4\pi }{6}}\end{array}$
  
  $\begin{array}{rl}\cos (x+y)& =\cos ({\displaystyle \frac{4\pi }{6}}+{\displaystyle \frac{3\pi }{6}})\\ & =\cos {\displaystyle \frac{7\pi }{6}}\\ & =\cos {\displaystyle \frac{\pi }{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{2}}\sqrt{3}}}}}\end{array}$

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