Himpunan penyelesaian dari 2 sin2 4x−1=0 untuk 0≤x≤2π adalah...

Pertanyaan

Himpunan penyelesaian dari $2\;\sin^2\;4x-1=0$ untuk $0\leq x\leq2\mathrm\pi$ adalah...

$\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac3{16}\mathrm\pi,\;\frac5{16}\mathrm\pi,\;\frac7{16}\mathrm\pi,\;\frac9{16}\mathrm\pi,\;\frac{11}{16}\mathrm\pi,\;\frac{13}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{19}{16}\mathrm\pi,\;\frac{21}{16}\mathrm\pi,\;\frac{23}{16}\mathrm\pi,\;\frac{25}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{29}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$
$\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac18\mathrm\pi,\;\frac5{16}\mathrm\pi,\;\frac7{16}\mathrm\pi,\;\frac9{16}\mathrm\pi,\;\frac34\mathrm\pi,\;\frac{13}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{19}{16}\mathrm\pi,\;\frac{21}{16}\mathrm\pi,\;\frac{23}{16}\mathrm\pi,\;\frac{25}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{29}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$
$\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac3{16}\mathrm\pi,\;\frac4{16}\mathrm\pi,\;\frac7{16}\mathrm\pi,\;\frac9{16}\mathrm\pi,\;\frac{10}{16}\mathrm\pi,\;\frac{13}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{18}{16}\mathrm\pi,\;\frac{21}{16}\mathrm\pi,\;\frac{23}{16}\mathrm\pi,\;\frac{25}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{29}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$
$\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac3{16}\mathrm\pi,\;\frac5{16}\mathrm\pi,\;\frac8{16}\mathrm\pi,\;\frac9{16}\mathrm\pi,\;\frac{11}{16}\mathrm\pi,\;\frac{12}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{19}{16}\mathrm\pi,\;\frac{20}{16}\mathrm\pi,\;\frac{23}{16}\mathrm\pi,\;\frac{24}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{29}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$
$\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac2{16}\mathrm\pi,\;\frac5{16}\mathrm\pi,\;\frac7{16}\mathrm\pi,\;\frac{10}{16}\mathrm\pi,\;\frac{11}{16}\mathrm\pi,\;\frac{13}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{18}{16}\mathrm\pi,\;\frac{21}{16}\mathrm\pi,\;\frac{24}{16}\mathrm\pi,\;\frac{25}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{30}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$

Y. Herlanda

Master Teacher

Mahasiswa/Alumni STKIP PGRI Jombang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah A.

Persamaan Trigonometri Dasar

Penyelesaian persamaan trigonometri dasar bentuk $\sin\;x=a$ dilakukan dengan cara: mengubah bentuk $\sin\;x=a$ menjadi $\sin\;x=\sin\;a$ .

Ingat persamaan trigonometri dasar berikut:

Jika $\sin\;x=\sin\;\alpha$ , nilai $x=\alpha+k\cdot2\mathrm\pi$ atau $x=\left(\mathrm\pi-\mathrm\alpha\right)+k\cdot2\mathrm\pi$ .

Diketahui $2\;\sin^2\;4x-1=0$ untuk $0\leq x\leq2\mathrm\pi$ , misal $\sin\;4x=p$ , maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space sin squared space 4 x minus 1 end cell equals 0 row cell 2 p squared minus 1 end cell equals 0 row cell 2 p squared end cell equals 1 row cell p squared end cell equals cell 1 half end cell row p equals cell plus-or-minus square root of 1 half end root end cell row blank equals cell plus-or-minus fraction numerator square root of 1 over denominator square root of 2 end fraction end cell row blank equals cell plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell plus-or-minus 1 half square root of 2 end cell row p equals cell 1 half square root of 2 space atau space p equals negative 1 half square root of 2 end cell end table$

Untuk $p=\frac12\sqrt2$ , maka:

$\begin{array}{rcl}\sin\;4x&=&\frac12\sqrt2\\\sin\;4x&=&\sin\;45^\circ\\\sin\;4x&=&\sin\;\frac14\mathrm\pi\\4x&=&\frac14\mathrm\pi+k\cdot2\mathrm\pi\\x&=&\frac1{16}\mathrm\pi+k\cdot\frac12\mathrm\pi\end{array}$

atau

$\begin{array}{rcl}\sin\;4x&=&\frac12\sqrt2\\\sin\;4x&=&\sin\;45^\circ\\\sin\;4x&=&\sin\;\frac14\mathrm\pi\\4x&=&\left(\mathrm\pi-\frac14\mathrm\pi\right)+k\cdot2\mathrm\pi\\4x&=&\frac34\mathrm\pi+k\cdot2\mathrm\pi\\x&=&\frac3{16}\mathrm\pi+k\cdot\frac12\mathrm\pi\end{array}$

Diperoleh

$\begin{array}{rcl}k&=&{0\rightarrow x=\frac1{16}\mathrm\pi+0\cdot\frac12\mathrm\pi=\frac1{16}\mathrm\pi\;\left(\mathrm M\right)\;\mathrm{atau}}\\k&=&{0\rightarrow x=\frac3{16}\mathrm\pi+0\cdot\frac12\mathrm\pi=\frac3{16}\mathrm\pi\;\left(\mathrm M\right)}\\k&=&{1\rightarrow x=\frac1{16}\mathrm\pi+1\cdot\frac12\mathrm\pi=\frac9{16}\mathrm\pi\;\left(\mathrm M\right)\;\mathrm{atau}}\\k&=&{1\rightarrow x=\frac3{16}\mathrm\pi+1\cdot\frac12\mathrm\pi=\frac{11}{16}\mathrm\pi\;\left(\mathrm M\right)}\\k&=&{2\rightarrow x=\frac1{16}\mathrm\pi+2\cdot\frac12\mathrm\pi=\frac{17}{16}\mathrm\pi\;\left(\mathrm M\right)\;\mathrm{atau}}\\k&=&{2\rightarrow x=\frac3{16}\mathrm\pi+2\cdot\frac12\mathrm\pi=\frac{19}{16}\mathrm\pi\;\left(\mathrm M\right)}\\k&=&{3\rightarrow x=\frac1{16}\mathrm\pi+3\cdot\frac12\mathrm\pi=\frac{25}{16}\mathrm\pi\;\left(\mathrm M\right)\;\mathrm{atau}}\\k&=&{3\rightarrow x=\frac3{16}\mathrm\pi+3\cdot\frac12\mathrm\pi=\frac{27}{16}\mathrm\pi\;\left(\mathrm M\right)}\end{array}$

Untuk $p=-\frac12\sqrt2$ , maka:

$\begin{array}{rcl}\sin\;4x&=&-\frac12\sqrt2\\\sin\;4x&=&\sin\;225^\circ\\\sin\;4x&=&\sin\;\frac54\mathrm\pi\\4x&=&\frac54\mathrm\pi+k\cdot2\mathrm\pi\\x&=&\frac5{16}\mathrm\pi+k\cdot\frac12\mathrm\pi\end{array}$

atau

$\begin{array}{rcl}\sin\;4x&=&-\frac12\sqrt2\\\sin\;4x&=&\sin\;225^\circ\\\sin\;4x&=&\sin\;\frac54\mathrm\pi\\4x&=&\left(\mathrm\pi-\frac54\mathrm\pi\right)+k\cdot2\mathrm\pi\\4x&=&-\frac14\mathrm\pi+k\cdot2\mathrm\pi\\x&=&-\frac1{16}\mathrm\pi+k\cdot\frac12\mathrm\pi\end{array}$

Diperoleh

$k=0\rightarrow x=\frac5{16}\mathrm\pi+0\cdot\frac12\mathrm\pi=\frac5{16}\mathrm\pi\;(\mathrm M)\;\mathrm{atau}\\k=0\rightarrow x=-\frac1{16}\mathrm\pi+0\cdot\frac12\mathrm\pi=-\frac1{16}\mathrm\pi\;(\mathrm{TM})\\k=1\rightarrow x=\frac5{16}\mathrm\pi+1\cdot\frac12\mathrm\pi=\frac{13}{16}\mathrm\pi\;(\mathrm M)\;\mathrm{atau}\\k=1\rightarrow x=-\frac1{16}\mathrm\pi+1\cdot\frac12\mathrm\pi=\frac7{16}\mathrm\pi\;(\mathrm M)\\k=2\rightarrow x=\frac5{16}\mathrm\pi+2\cdot\frac12\mathrm\pi=\frac{21}{16}\mathrm\pi\;(\mathrm M)\;\mathrm{atau}\\k=2\rightarrow x=-\frac1{16}\mathrm\pi+2\cdot\frac12\mathrm\pi=\frac{31}{16}\mathrm\pi\;(\mathrm M)\\k=3\rightarrow x=\frac5{16}\mathrm\pi+3\cdot\frac12\mathrm\pi=\frac{29}{16}\mathrm\pi\;(\mathrm M)\;\mathrm{atau}\\k=3\rightarrow x=-\frac1{16}\mathrm\pi+3\cdot\frac12\mathrm\pi=\frac{23}{16}\mathrm\pi\;(\mathrm M)$

Keterangan:

M (Memenuhi), TM (Tidak Memenuhi)

Dengan demikian, himpunan penyelesaiannya adalah $\style{font-size:12px}{\left\{\frac1{16}\mathrm\pi,\;\frac3{16}\mathrm\pi,\;\frac5{16}\mathrm\pi,\;\frac7{16}\mathrm\pi,\;\frac9{16}\mathrm\pi,\;\frac{11}{16}\mathrm\pi,\;\frac{13}{16}\mathrm\pi,\;\frac{15}{16}\mathrm\pi,\;\right\}\\\left\{\frac{17}{16}\mathrm\pi,\;\frac{19}{16}\mathrm\pi,\;\frac{21}{16}\mathrm\pi,\;\frac{23}{16}\mathrm\pi,\;\frac{25}{16}\mathrm\pi,\;\frac{27}{16}\mathrm\pi,\;\frac{29}{16}\mathrm\pi,\;\frac{31}{16}\mathrm\pi\right\}}$ .

Oleh karena itu, jawaban yang benar adalah A.