B. 2 Persamaan Trigonometri Bentuk Kuadrat
Persamaan trigonometri terkadang juga terdapat dalam bentuk kuadrat, sehingga penyelesaiannya menyesuaikan dengan persamaan kuadrat tersebut yaitu proses faktorisasi, atau melengkapkan kudrat sempurna,dan atau dengan rumus ABC.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\sin x-2\sin ^{2}x=0\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &\sin x-2\sin ^{2}x=0\: \: (\textbf{lalu difaktorkan})\\ &\sin x\left ( 1-2\sin x \right )=0\\ &\sin x=0\: \: \textrm{atau}\: \: 1-2\sin x=0\\ &\textrm{selanjutnya}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\sin x&=0\\ \sin x&=\sin 0^{\circ}\\ x&=0^{\circ}+k.360^{\circ}\\ &\color{red}\textrm{atau}\\ x&=180^{\circ}+k.360^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=0^{\circ}\: \: \textrm{dan}\: \: 180^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=360^{\circ}\: \: \textrm{dan}\: \: \color{red}540^{\circ} \end{aligned}&\begin{aligned}\sin x&=\displaystyle \frac{1}{2}\\ \sin x&=\sin 30^{\circ}\\ x&=30^{\circ}+k.360^{\circ}\\ &\color{red}\textrm{atau}\\ x&=\left (180^{\circ}-30^{\circ} \right )+k.360^{\circ}\\ &=150^{\circ}+k.360^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=30^{\circ}\: \: \textrm{dan}\: \: 150^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=\color{red}390^{\circ}\: \: \color{black}\textrm{dan}\: \: \color{red}510^{\circ} \end{aligned} \\\hline \end{array}\\ &\textbf{HP}=\left \{ 0^{\circ},30^{\circ},150^{\circ},180^{\circ},360^{\circ} \right \} \end{array}$
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &2\tan ^{2}\theta -\sec \theta +1=0\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &\tan ^{2}\theta -\sec \theta +1=0\\ &2\left (\sec ^{2}\theta -1 \right )-\sec \theta +1=0\\ &2\sec ^{2}\theta -\sec \theta -1=0\: \: (\textbf{lalu difaktorkan})\\ &\left (2\sec \theta +1 \right )\left ( \sec \theta -1 \right )=0\\ &\left (2\sec \theta +1 \right )=0\: \: \textrm{atau}\: \: \left (\sec \theta -1 \right )=0\\ &\sec \theta =-\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \sec \theta =1\\ &\displaystyle \frac{1}{\cos \theta }=-\frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{1}{\cos \theta }=1\\ &\cos \theta =-2\: (\textbf{tidak mungkin})\: \: \textrm{atau}\: \: \cos \theta =1\\ &\textrm{selanjutnya}\\ &\cos \theta =1\Leftrightarrow \cos \theta =\cos 0^{\circ}\\ &\Leftrightarrow \theta =\pm 0^{\circ}+k.360^{\circ}\Leftrightarrow \theta =k.360^{\circ}\\ &k=0\Rightarrow x=0^{\circ}\\ &k=1\Rightarrow x=360^{\circ}\\ &k=2\Rightarrow x=\color{red}720^{\circ}\: \: \textrm{tidak memenuhi}\\ &\textbf{HP}=\left \{ 0^{\circ},360^{\circ} \right \} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &5\cos ^{2}\beta +3\cos \beta =2\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \beta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &5\cos ^{2}\beta +3\cos \beta =2\\ &5\cos ^{2}\beta +3\cos \beta -2=0\: \: (\textbf{lalu difaktorkan})\\ &\left ( 5\cos \beta -2 \right )\left ( \cos \beta +1 \right )=0\\ &\left ( 5\cos \beta -2 \right )=0\: \: \textrm{atau}\: \: \left ( \cos \beta +1 \right )=0\\ &5\cos \beta -2=0\: \: \textrm{atau}\: \: \cos \beta +1=0\\ &\cos \beta =\displaystyle \frac{2}{5}\: \: \textrm{atau}\: \: \cos \beta =-1\\ &\cos \beta =\cos 66,4^{\circ}\: \: \textrm{atau}\: \: \cos \beta =180^{\circ}\\ &\textrm{selanjutnya}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\beta &=\pm 66,4^{\circ}+k.360^{\circ}\\ k=0&\Rightarrow \beta =66,4^{\circ}\: \: \textrm{atau}\\ &\beta =-66,4^{\circ}\: \: (\textrm{tm})\\ k=1&\Rightarrow \beta =426,4^{\circ}\: \: (\textrm{tm})\\ & \textrm{atau}\: \: \beta =293,6^{\circ}\\ k=2&\Rightarrow \beta \: \: \textrm{tidak ada }\\ &\qquad\textrm{yang memenuhi} \end{aligned}&\begin{aligned}\beta &=\pm 180^{\circ}+k.360^{\circ}\\ k=0&\Rightarrow \beta =180^{\circ}\: \: \textrm{atau}\\ &\beta =-180^{\circ}\: \: (\textrm{tm})\\ k=1&\Rightarrow \beta =540^{\circ}\: \: (\textrm{tm})\\ &\textrm{atau}\: \: \beta =180^{\circ}\\ k=2&\Rightarrow \beta \: \: \textrm{tidak ada }\\ &\qquad\textrm{yang memenuhi} \end{aligned} \\\hline \end{array}\\ &\textbf{HP}=\left \{ 66,4^{\circ},180^{\circ},293,6^{\circ} \right \} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &2\sin ^{2}\gamma +3\cos \gamma =3\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \gamma \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &2\sin ^{2}\gamma +3\cos \gamma =3\\ &2\left ( 1-\cos ^{2}\gamma \right ) +3\cos \gamma -3=0\\ &2-2\cos ^{2}\gamma +3\cos \gamma -3=0\\ &-2\cos ^{2}\gamma +3\cos \gamma -1=0\\ &2\cos ^{2}\gamma -3\cos \gamma +1=0\: \: (\textbf{lalu difaktorkan})\\ &\left ( 2\cos \gamma -1 \right )\left ( \cos \gamma -1 \right )=0\\ &\left ( 2\cos \gamma -1 \right )=0\: \:\textrm{ atau}\: \: \left ( \cos \gamma -1 \right )=0\\ &\cos \gamma =\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \cos \gamma =1\\ &\cos \gamma =\cos 60^{\circ}\: \: \textrm{atau}\: \: \cos \gamma =0^{\circ}\\ &\textrm{selanjutnya}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\gamma &=\pm 60^{\circ}+k.360^{\circ}\\ k=0&\Rightarrow \gamma =60^{\circ}\: \: \textrm{atau}\\ &\gamma =-60^{\circ}\: \: (\textrm{tm})\\ k=1&\Rightarrow \gamma =420^{\circ}\: \: (\textrm{tm})\\ & \textrm{atau}\: \: \gamma =300^{\circ}\\ k=2&\Rightarrow \gamma \: \: \textrm{tidak ada }\\ &\quad\textrm{yang memenuhi} \end{aligned}&\begin{aligned}\gamma &=\pm 0^{\circ}+k.360^{\circ}\\ \gamma &=0^{\circ}+k.360^{\circ}\\ k=0&\Rightarrow \gamma =0^{\circ}\\ k=1&\Rightarrow \gamma =360^{\circ}\\ k=2&\Rightarrow \gamma \: \: \textrm{tidak ada}\\ &\quad\textrm{yang memenuhi}\\ &\vdots \\ \end{aligned} \\\hline \end{array}\\ &\textbf{HP}=\left \{ 0^{\circ},60^{\circ},300^{\circ},360^{\circ} \right \} \end{array}$
B. 3 Persamaan Trigonometri Bentuk a sin x + b cos x
Selain bentuk sederhana seperti yang telah diuraikan pada materi sebelumnya (lihat di sini), terdapat persamaan trigonometri bentuk $a\sin x+b\cos x$. Bentuk $a\sin x+b\cos x$ ini dalam penyelesaiannya diubah ke dalam bentuk $k\cos (x-\alpha )$. Adapun untuk menemukan pembuktian dari kesamaan rumus ini, Anda harus mempelajari materi rumus trigonometri jumlah dan selisih dua sudut.
$\begin{aligned}a\sin x&+b\cos x=k\cos \left ( x-\theta \right )\\ \color{purple}\textrm{denga}&\color{purple}\textrm{n}:\: \: \\ &k=\sqrt{a^{2}+b^{2}}\\ &\tan \theta =\displaystyle \frac{a}{b}\\ &(a>0\: \: \textrm{dan}\: \: b>0,\: \textrm{maka}\: \theta \: \textrm{di kuadran I})\\ &(a>0\: \: \textrm{dan}\: \: b<0,\: \textrm{maka}\: \theta \: \textrm{di kuadran II})\\ &(a<0\: \: \textrm{dan}\: \: b<0,\: \textrm{maka}\: \theta \: \textrm{di kuadran III})\\ &(a<0\: \: \textrm{dan}\: \: b>0,\: \textrm{maka}\: \theta \: \textrm{di kuadran IV})\\\\ &\textrm{dengan}\: \: a\: \: \textrm{pada sumbu Y dan}\\ &\: \: \, \quad\quad\quad b\: \: \textrm{pada sumbu X} \end{aligned}$
$\begin{aligned}&\textbf{Dan ingat juga tabel nilai tangen}\\ &\textbf{berikut}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \theta &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\\ &&&&&\\\hline \tan \theta &\color{blue}0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textbf{TD}\\ &&&&&\\\hline \theta &120^{\circ}&135^{\circ}&150^{\circ}&180^{\circ}&\\ &&&&&\\\hline \tan \theta &-\sqrt{3}&-1&-\displaystyle \frac{1}{3}\sqrt{3}&\color{blue}0&\\ &&&&&\\\hline \end{array} \end{aligned}$.
$\begin{aligned}&\begin{array}{|c|c|c|c|c|c|}\hline \theta &180^{\circ}&210^{\circ}&225^{\circ}&240^{\circ}&270^{\circ}\\ &&&&&\\\hline \tan \theta &\color{blue}0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textbf{TD}\\ &&&&&\\\hline \theta &300^{\circ}&315^{\circ}&345^{\circ}&360^{\circ}&\\ &&&&&\\\hline \tan \theta &-\sqrt{3}&-1&-\displaystyle \frac{1}{3}\sqrt{3}&\color{blue}0&\\ &&&&&\\\hline \end{array} \end{aligned}$.
Untuk lebih lanjut tentang bukti dan lain sebagainya akan dipelajari di subbab berikutnya setelah materi persamaan trigonometri ini.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\sin x+\sqrt{3}\cos x=2\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &\sin x+\sqrt{3}\cos x=2\quad \left (\textbf{ingat}:a=1,\: b=\sqrt{3} \right )\\ &\sin x+\sqrt{3}\cos x=k\cos \left ( x-\theta \right )=2\\ &\begin{cases} k & =\sqrt{1^{2}+\left ( \sqrt{3} \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{\sqrt{3}}=\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta =30^{\circ} \end{cases}\\ &\qquad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a,b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x+\sqrt{3}\cos x=k\cos \left ( x-\theta \right )=2\\ &\Leftrightarrow 2\cos\left ( x-30^{\circ} \right )=2\\ &\Leftrightarrow \: \: \, \cos \left ( x-30^{\circ} \right )=1\\ &\Leftrightarrow \: \: \,\cos \left ( x-30^{\circ} \right )=\cos 0^{\circ}\\ &\Leftrightarrow \quad x-30^{\circ} =\pm 0^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=30^{\circ}\pm 0^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=30^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x=30^{\circ}\qquad (\color{blue}\textrm{memenuhi})\\ &k=1\Rightarrow x=390^{\circ}\qquad (\color{red}\textrm{tm})\\ \end{aligned}\\ &\textbf{HP}=\left \{30^{\circ} \right \} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\sin x-\sqrt{3}\cos x=\sqrt{2}\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &\sin x-\sqrt{3}\cos x=\sqrt{2}\: \: \left (\textbf{ingat}:a=1,\: b=-\sqrt{3} \right )\\ &\sin x-\sqrt{3}\cos x=k\cos \left ( x-\theta \right )=\sqrt{2}\\ &\begin{cases} k & =\sqrt{1^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{-\sqrt{3}}=-\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta =150^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran II, karena}\: \: a>0,\: b<0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x-\sqrt{3}\cos x=k\cos \left ( x-\theta \right )=\sqrt{2}\\ &\Leftrightarrow 2\cos\left ( x-150^{\circ} \right )=\sqrt{2}\\ &\Leftrightarrow \: \: \, \cos \left ( x-150^{\circ} \right )=\displaystyle \frac{\sqrt{2}}{2}=\displaystyle \frac{1}{2}\sqrt{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-150^{\circ} \right )=\cos 45^{\circ}\\ &\Leftrightarrow \quad x-150^{\circ} =\pm 45^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=150^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x=150^{\circ}+45^{\circ}=195^{\circ}\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x=150^{\circ}-45^{\circ}=105^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=150^{\circ}\pm 45^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{105^{\circ},195^{\circ} \right \} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\sqrt{6}\sin x+\sqrt{2}\cos x=2\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &\sqrt{6}\sin x+\sqrt{2}\cos x=2\: \: \left (\textbf{ingat}:a=\sqrt{6},\: b=\sqrt{2} \right )\\ &\sqrt{6}\sin x+\sqrt{2}\cos x=k\cos \left ( x-\theta \right )=2\\ &\begin{cases} k & =\sqrt{\left (\sqrt{6} \right )^{2}+\left ( \sqrt{2} \right )^{2}}=\sqrt{8}=2\sqrt{2} \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\Rightarrow \theta =60^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a>0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sqrt{6}\sin x+\sqrt{2}\cos x=k\cos \left ( x-\theta \right )=2\\ &\Leftrightarrow 2\sqrt{2}\cos\left ( x-60^{\circ} \right )=2\\ &\Leftrightarrow \: \: \, \cos \left ( x-60^{\circ} \right )=\displaystyle \frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}=\displaystyle \frac{1}{2}\sqrt{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-60^{\circ} \right )=\cos 45^{\circ}\\ &\Leftrightarrow \quad x-60^{\circ} =\pm 45^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=60^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x=60^{\circ}+45^{\circ}=105^{\circ}\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x=60^{\circ}-45^{\circ}=15^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=60^{\circ}\pm 45^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{15^{\circ},105^{\circ} \right \} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\cos x-\sqrt{3}\sin x=1\: \: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 360^{\circ}\\\\ &\textrm{Jawab}:\\ &-\sqrt{3}\sin x+\cos x=1\: \: \left (\textbf{ingat}:a=-\sqrt{3},\: b=1 \right )\\ &-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\begin{cases} k & =\sqrt{\left ( -\sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta =300^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran IV, karena}\: \: a<0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\Leftrightarrow 2\cos\left ( x-300^{\circ} \right )=1\\ &\Leftrightarrow \: \: \, \cos \left ( x-300^{\circ} \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-300^{\circ} \right )=\cos 60^{\circ}\\ &\Leftrightarrow \quad x-300^{\circ} =\pm 60^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=300^{\circ}\pm 60^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x=300^{\circ}+60^{\circ}=360^{\circ}=0^{\circ}\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x=300^{\circ}-60^{\circ}=240^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=300^{\circ}\pm 60^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{0^{\circ},240^{\circ},360^{\circ} \right \} \end{array}$
DAFTAR PUSTAKA
- Kanginan, M., Nurdiasyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
- Noormandiri, B. K. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
- Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
- Sukino. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Tidak ada komentar:
Posting Komentar
Informasi