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:
$\begin{array}{rl}\cos \left({\displaystyle \frac{\pi }{3}}x\right)& =0\\ {\displaystyle \frac{\pi }{3}}x& =\{-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}},\cdots \}\\ x& =\{-{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{9}{2}},\cdots \}\\ x& ={\displaystyle \frac{3}{2}}\end{array}$

$\begin{array}{rl}\cos \left({\displaystyle \frac{\pi }{2}}x\right)& =0\\ {\displaystyle \frac{\pi }{2}}x& =\{-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}},{\displaystyle \frac{5\pi }{2}},\cdots \}\\ x& =\{-1,1,3,5,\cdots \}\\ x& =1,3\end{array}$

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:
$\begin{array}{rl}\cos 2x+\cos 3x& =0\\ \cos 2x& =-\cos 3x\\ \cos 2x& =\cos (\pi -3x)\end{array}$
$\begin{array}{rlrlrl}2x& =\pi -3x+2k\pi & \text{atau}& 2x& =& -(\pi -3x)+2k\pi \\ 5x& =\pi +2k\pi & \text{atau}& 2x& =& -\pi +3x+2k\pi \\ x& ={\displaystyle \frac{\pi }{5}}+{\displaystyle \frac{2}{5}}k\pi & \text{atau}& -x& =& -\pi +2k\pi \\ x& ={\displaystyle \frac{\pi }{5}}+{\displaystyle \frac{2k\pi }{5}}& \text{atau}& x& =& \pi -2k\pi \\ x& =\{{\displaystyle \frac{\pi }{5}},{\displaystyle \frac{3\pi }{5}},\pi ,{\displaystyle \frac{7\pi }{5}},{\displaystyle \frac{9\pi }{5}}\}& \text{atau}& x& =& \{\pi \}\end{array}$

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:
$\begin{array}{rl}\tan 2x& ={\displaystyle \frac{1}{3}}\sqrt{3}\\ \tan 2x& =\tan {\displaystyle \frac{1}{6}}\pi \\ 2x& =\{{\displaystyle \frac{1}{6}}\pi ,{\displaystyle \frac{7}{6}}\pi ,{\displaystyle \frac{13}{6}}\pi ,{\displaystyle \frac{19}{6}}\pi \}\\ x& =\{{\displaystyle \frac{1}{12}}\pi ,{\displaystyle \frac{7}{12}}\pi ,{\displaystyle \frac{13}{12}}\pi ,{\displaystyle \frac{19}{12}}\pi \}\\ & =\left\{{\displaystyle \frac{13}{12}}\pi \right\}\end{array}$

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