Exercise Zone : Pertidaksamaan Trigonometri

Berikut ini adalah kumpulan soal mengenai pertidaksamaan trigonometri tingkat dasar. Mau tanya soal? gabung aja ke grup Facebook atau Telegram.

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

4. 1\leq x\leq2 atau 3\leq x\leq 4
5. 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,

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No. 3

Himpunan penyelesaian pertidaksamaan \tan2x\leq\dfrac13\sqrt3 dengan \pi\leq x\leq\dfrac32\pi adalah

1. \left\{x\mid \pi\leq x\leq\dfrac76\pi\text{ atau }\dfrac54\pi\leq x\leq\dfrac32\pi\right\}
2. \left\{x\mid \pi\leq x\leq\dfrac76\pi\text{ atau }\dfrac54\pi\lt x\leq\dfrac32\pi\right\}
3. \left\{x\mid \pi\leq x\lt\dfrac54\pi\text{ atau }\dfrac54\pi\lt x\leq\dfrac32\pi\right\}

4. \left\{x\mid \dfrac76\pi\leq x\lt\dfrac32\pi\right\}
5. \left\{x\mid \dfrac76\pi\leq x\leq\dfrac32\pi\right\}

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Pembuat nol:
\(\eqalign{ \tan2x&=\dfrac13\sqrt3\\ \tan2x&=\tan\dfrac16\pi\\ 2x&=\left\{\dfrac16\pi,\dfrac76\pi,\dfrac{13}6\pi,\dfrac{19}6\pi\right\}\\ x&=\left\{\dfrac1{12}\pi,\dfrac7{12}\pi,\dfrac{13}{12}\pi,\dfrac{19}{12}\pi\right\}\\ &=\left\{\dfrac{13}{12}\pi\right\} }\)

Jadi, \left\{x\mid \pi\leq x\leq\dfrac{13}{12}\pi\text{ atau }\dfrac54\pi\lt x\leq\dfrac32\pi\right\}