Pertidaksamaan Trigonometri : Soal dan Pembahasan (DISCONTINUED)

Berikut adalah kumpulan soal mengenai pertidaksamaan trigonometri. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Jika 0\leq x\leq4, maka nilai x yang memenuhi pertidaksamaan \cos\left(\dfrac{\pi}3x\right)\cdot\cos\left(\dfrac{\pi}2x\right)\leq0 adalah....

1\leq x\leq\dfrac32 atau 3\leq x\leq 4	1\leq x\leq\dfrac32 atau 2\leq x\leq \dfrac52
1\leq x\leq\dfrac32 atau 2\leq x\leq 4	1\leq x\leq2 atau 3\leq x\leq 4
1\leq x\leq\dfrac32 atau 2\leq x\leq 3

Pembuat nol:
\begin{aligned}
\cos\left(\dfrac{\pi}3x\right)&=0\\
\dfrac{\pi}3x&=\{-\dfrac{\pi}2,\dfrac{\pi}2,\dfrac{3\pi}2,\cdots\}\\
x&=\{-\dfrac32,\dfrac32,\dfrac92,\cdots\}\\
x&={\dfrac32}
\end{aligned}

\begin{aligned}
\cos\left(\dfrac{\pi}2x\right)&=0\\
\dfrac{\pi}2x&=\{-\dfrac{\pi}2,\dfrac{\pi}2,\dfrac{3\pi}2,\dfrac{5\pi}2,\cdots\}\\
x&=\{-1,1,3,5,\cdots\}\\
x&={1,3}
\end{aligned}

1\leq x\leq\dfrac32 atau 3\leq x\leq4

Jadi, 1\leq x\leq\dfrac32 atau 3\leq x\leq 4.
JAWAB: A

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No. 2

Fungsi f(x)=\cos2x+\cos3x, 0\lt x\lt2\pi akan terletak di bawah sumbu x pada interval

1. \dfrac{3\pi}5\lt x\lt\dfrac{7\pi}5
2. \dfrac{\pi}5\lt x\lt\pi
3. 0\lt x\lt\dfrac{7\pi}5

4. \dfrac{7\pi}5\lt x\lt\dfrac{9\pi}5
5. \pi\lt x\lt\dfrac{7\pi}5

Pembatas:
\begin{aligned}
\cos2x+\cos3x&=0\\
\cos2x&=-\cos3x\\
\cos2x&=\cos(\pi-3x)
\end{aligned}
\begin{aligned}
2x&=\pi-3x+2k\pi\qquad&{\color{black}\text{atau}}\qquad&2x&=&-(\pi-3x)+2k\pi\\
5x&=\pi+2k\pi\qquad&{\color{black}\text{atau}}\qquad&2x&=&-\pi+3x+2k\pi\\
x&=\dfrac\pi5+\dfrac25k\pi\qquad&{\color{black}\text{atau}}\qquad&-x&=&-\pi+2k\pi\\
x&=\dfrac\pi5+\dfrac{2k\pi}5\qquad&{\color{black}\text{atau}}\qquad&x&=&\pi-2k\pi\\
x&=\left\{\dfrac\pi5,\dfrac{3\pi}5,\pi,\dfrac{7\pi}5,\dfrac{9\pi}5\right\}\qquad&{\color{black}\text{atau}}\qquad&x&=&\{\pi\}
\end{aligned}

Jadi,