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Nilai x yang memenuhi persamaan cos 2x-sin x=0 untuk 0°<x<360°

Trigonometri Matematika · SMA medium

Nilai xx yang memenuhi persamaan cos⁡2x−sin⁡x=0\cos 2x-\sin x=0 untuk 0∘<x<360∘0^\circ<x<360^\circ adalah ....

  1. A
    {30∘,150∘}\{30^\circ,150^\circ\}
  2. B
    {30∘,270∘}\{30^\circ,270^\circ\}
  3. C
    {30∘,150∘,180∘}\{30^\circ,150^\circ,180^\circ\}
  4. D
    {60∘,120∘,300∘}\{60^\circ,120^\circ,300^\circ\}
  5. E
    {30∘,150∘,270∘}\{30^\circ,150^\circ,270^\circ\}
    ✓

Kunci jawaban

E

Penjelasan singkat

cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x sehingga 2sin⁡2x+sin⁡x−1=0⇒(2sin⁡x−1)(sin⁡x+1)=02\sin^2x+\sin x-1=0\Rightarrow(2\sin x-1)(\sin x+1)=0. Maka sin⁡x=12\sin x=\tfrac12 (x=30∘,150∘x=30^\circ,150^\circ) atau sin⁡x=−1\sin x=-1 (x=270∘x=270^\circ).

Gunakan cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x:

1−2sin⁡2x−sin⁡x=0⇒2sin⁡2x+sin⁡x−1=0⇒(2sin⁡x−1)(sin⁡x+1)=01-2\sin^2x-\sin x=0 \Rightarrow 2\sin^2x+\sin x-1=0 \Rightarrow (2\sin x-1)(\sin x+1)=0.

  • sin⁡x=12⇒x=30∘, 150∘\sin x=\tfrac12 \Rightarrow x=30^\circ,\,150^\circ.
  • sin⁡x=−1⇒x=270∘\sin x=-1 \Rightarrow x=270^\circ.

Himpunan: {30∘,150∘,270∘}\{30^\circ,150^\circ,270^\circ\}. Jawabannya E.

⚡ Latihan Matematika SMA