Lanjutan 4 Persamaan Trigonometri

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.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x-2{\sin }^{2}x=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x-2{\sin }^{2}x=0(\mathbf{\text{lalu difaktorkan}})\\ & \sin x(1-2\sin x)=0\\ & \sin x=0\text{atau}1-2\sin x=0\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\sin x& =0\\ \sin x& =\sin 0^{\circ }\\ x& =0^{\circ }+k{.360}^{\circ }\\ & \text{atau}\\ x& ={180}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x& =0^{\circ }\text{dan}{180}^{\circ }\\ \text{saat}& k=1\\ x& ={360}^{\circ }\text{dan}{540}^{\circ }\end{array}& \begin{array}{rl}\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& ={30}^{\circ }+k{.360}^{\circ }\\ & \text{atau}\\ x& =({180}^{\circ }-{30}^{\circ })+k{.360}^{\circ }\\ & ={150}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\text{dan}{150}^{\circ }\\ \text{saat}& k=1\\ x& ={390}^{\circ }\text{dan}{510}^{\circ }\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{30}^{\circ },{150}^{\circ },{180}^{\circ },{360}^{\circ }\}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 2{\tan }^2\theta -\sec \theta +1=0\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & {\tan }^2\theta -\sec \theta +1=0\\ & 2({\sec }^2\theta -1)-\sec \theta +1=0\\ & 2{\sec }^2\theta -\sec \theta -1=0(\mathbf{\text{lalu difaktorkan}})\\ & (2\sec \theta +1)(\sec \theta -1)=0\\ & (2\sec \theta +1)=0\text{atau}(\sec \theta -1)=0\\ & \sec \theta =-{\displaystyle \frac{1}{2}\text{atau}\sec \theta =1}\\ & {\displaystyle \frac{1}{\cos \theta }=-\frac{1}{2}\text{atau}{\displaystyle \frac{1}{\cos \theta }=1}}\\ & \cos \theta =-2(\mathbf{\text{tidak mungkin}})\text{atau}\cos \theta =1\\ & \text{selanjutnya}\\ & \cos \theta =1\iff \cos \theta =\cos 0^{\circ }\\ & \iff \theta =\pm 0^{\circ }+k{.360}^{\circ }\iff \theta =k{.360}^{\circ }\\ & k=0\Rightarrow x=0^{\circ }\\ & k=1\Rightarrow x={360}^{\circ }\\ & k=2\Rightarrow x={720}^{\circ }\text{tidak memenuhi}\\ & \mathbf{\text{HP}}=\{0^{\circ },{360}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 5{\cos }^2\beta +3\cos \beta =2\text{untuk}{0}^{\circ }\le \beta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & 5{\cos }^2\beta +3\cos \beta =2\\ & 5{\cos }^2\beta +3\cos \beta -2=0(\mathbf{\text{lalu difaktorkan}})\\ & (5\cos \beta -2)(\cos \beta +1)=0\\ & (5\cos \beta -2)=0\text{atau}(\cos \beta +1)=0\\ & 5\cos \beta -2=0\text{atau}\cos \beta +1=0\\ & \cos \beta ={\displaystyle \frac{2}{5}\text{atau}\cos \beta =-1}\\ & \cos \beta =\cos 66,4^{\circ }\text{atau}\cos \beta ={180}^{\circ }\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\beta & =\pm 66,4^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \beta =66,4^{\circ }\text{atau}\\ & \beta =-66,4^{\circ }(\text{tm})\\ k=1& \Rightarrow \beta =426,4^{\circ }(\text{tm})\\ & \text{atau}\beta =293,6^{\circ }\\ k=2& \Rightarrow \beta \text{tidak ada }\\ & \text{yang memenuhi}\end{array}& \begin{array}{rl}\beta & =\pm {180}^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \beta ={180}^{\circ }\text{atau}\\ & \beta =-{180}^{\circ }(\text{tm})\\ k=1& \Rightarrow \beta ={540}^{\circ }(\text{tm})\\ & \text{atau}\beta ={180}^{\circ }\\ k=2& \Rightarrow \beta \text{tidak ada }\\ & \text{yang memenuhi}\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{66,4^{\circ },{180}^{\circ },293,6^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 2{\sin }^2\gamma +3\cos \gamma =3\text{untuk}{0}^{\circ }\le \gamma \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & 2{\sin }^2\gamma +3\cos \gamma =3\\ & 2(1-{\cos }^2\gamma )+3\cos \gamma -3=0\\ & 2-2{\cos }^2\gamma +3\cos \gamma -3=0\\ & -2{\cos }^2\gamma +3\cos \gamma -1=0\\ & 2{\cos }^2\gamma -3\cos \gamma +1=0(\mathbf{\text{lalu difaktorkan}})\\ & (2\cos \gamma -1)(\cos \gamma -1)=0\\ & (2\cos \gamma -1)=0\text{ atau}(\cos \gamma -1)=0\\ & \cos \gamma ={\displaystyle \frac{1}{2}\text{atau}\cos \gamma =1}\\ & \cos \gamma =\cos {60}^{\circ }\text{atau}\cos \gamma =0^{\circ }\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\gamma & =\pm {60}^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \gamma ={60}^{\circ }\text{atau}\\ & \gamma =-{60}^{\circ }(\text{tm})\\ k=1& \Rightarrow \gamma ={420}^{\circ }(\text{tm})\\ & \text{atau}\gamma ={300}^{\circ }\\ k=2& \Rightarrow \gamma \text{tidak ada }\\ & \text{yang memenuhi}\end{array}& \begin{array}{rl}\gamma & =\pm 0^{\circ }+k{.360}^{\circ }\\ \gamma & =0^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \gamma =0^{\circ }\\ k=1& \Rightarrow \gamma ={360}^{\circ }\\ k=2& \Rightarrow \gamma \text{tidak ada}\\ & \text{yang memenuhi}\\ & ⋮\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{60}^{\circ },{300}^{\circ },{360}^{\circ }\}\end{array}$

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.

$\begin{array}{rl}a\sin x& +b\cos x=k\cos (x-\theta )\\ \text{denga}& \text{n}:\\ & k=\sqrt{a^2+b^2}\\ & \tan \theta ={\displaystyle \frac{a}{b}}\\ & (a>0\text{dan}b>0,\text{maka}\theta \text{di kuadran I})\\ & (a>0\text{dan}b<0,\text{maka}\theta \text{di kuadran II})\\ & (a<0\text{dan}b<0,\text{maka}\theta \text{di kuadran III})\\ & (a<0\text{dan}b>0,\text{maka}\theta \text{di kuadran IV})\\ \\ & \text{dengan}a\text{pada sumbu Y dan}\\ & b\text{pada sumbu X}\end{array}$

$\begin{array}{rl}& \mathbf{\text{Dan ingat juga tabel nilai tangen}}\\ & \mathbf{\text{berikut}}\\ & \begin{array}{|cccccc|}\hline \theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }\\ & & & & & \\ \tan \theta & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \mathbf{\text{TD}}\\ & & & & & \\ \theta & {120}^{\circ }& {135}^{\circ }& {150}^{\circ }& {180}^{\circ }& \\ & & & & & \\ \tan \theta & -\sqrt{3}& -1& -{\displaystyle \frac{1}{3}\sqrt{3}}& 0& \\ & & & & & \\ \hline\end{array}\end{array}$.

$\begin{array}{rl}& \begin{array}{|cccccc|}\hline \theta & {180}^{\circ }& {210}^{\circ }& {225}^{\circ }& {240}^{\circ }& {270}^{\circ }\\ & & & & & \\ \tan \theta & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \mathbf{\text{TD}}\\ & & & & & \\ \theta & {300}^{\circ }& {315}^{\circ }& {345}^{\circ }& {360}^{\circ }& \\ & & & & & \\ \tan \theta & -\sqrt{3}& -1& -{\displaystyle \frac{1}{3}\sqrt{3}}& 0& \\ & & & & & \\ \hline\end{array}\end{array}$.

Untuk lebih lanjut tentang bukti dan lain sebagainya akan dipelajari di subbab berikutnya setelah materi persamaan trigonometri ini.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x+\sqrt{3}\cos x=2\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x+\sqrt{3}\cos x=2(\mathbf{\text{ingat}}:a=1,b=\sqrt{3})\\ & \sin x+\sqrt{3}\cos x=k\cos (x-\theta )=2\\ & \{\begin{array}{ll}k& =\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{\sqrt{3}}={\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta ={30}^{\circ }}}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x+\sqrt{3}\cos x=k\cos (x-\theta )=2\\ & \iff 2\cos (x-{30}^{\circ })=2\\ & \iff \cos (x-{30}^{\circ })=1\\ & \iff \cos (x-{30}^{\circ })=\cos 0^{\circ }\\ & \iff x-{30}^{\circ }=\pm 0^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm 0^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={30}^{\circ }(\text{memenuhi})\\ & k=1\Rightarrow x={390}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\left\{{30}^{\circ }\right\}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x-\sqrt{3}\cos x=\sqrt{2}\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x-\sqrt{3}\cos x=\sqrt{2}(\mathbf{\text{ingat}}:a=1,b=-\sqrt{3})\\ & \sin x-\sqrt{3}\cos x=k\cos (x-\theta )=\sqrt{2}\\ & \{\begin{array}{ll}k& =\sqrt{1^2+{(-\sqrt{3})}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{-\sqrt{3}}=-{\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta ={150}^{\circ }}}}\end{array}\\ & \text{sudut}\theta \text{di kuadran II, karena}a>0,b<0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x-\sqrt{3}\cos x=k\cos (x-\theta )=\sqrt{2}\\ & \iff 2\cos (x-{150}^{\circ })=\sqrt{2}\\ & \iff \cos (x-{150}^{\circ })={\displaystyle \frac{\sqrt{2}}{2}={\displaystyle \frac{1}{2}\sqrt{2}}}\\ & \iff \cos (x-{150}^{\circ })=\cos {45}^{\circ }\\ & \iff x-{150}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={150}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={150}^{\circ }+{45}^{\circ }={195}^{\circ }(\text{mm})\text{atau}\\ & x={150}^{\circ }-{45}^{\circ }={105}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={150}^{\circ }\pm {45}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{{105}^{\circ },{195}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=2\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=2(\mathbf{\text{ingat}}:a=\sqrt{6},b=\sqrt{2})\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=k\cos (x-\theta )=2\\ & \{\begin{array}{ll}k& =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(\sqrt{2}\right)}^2}=\sqrt{8}=2\sqrt{2}\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a>0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sqrt{6}\sin x+\sqrt{2}\cos x=k\cos (x-\theta )=2\\ & \iff 2\sqrt{2}\cos (x-{60}^{\circ })=2\\ & \iff \cos (x-{60}^{\circ })={\displaystyle \frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}={\displaystyle \frac{1}{2}\sqrt{2}}}\\ & \iff \cos (x-{60}^{\circ })=\cos {45}^{\circ }\\ & \iff x-{60}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={60}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={60}^{\circ }+{45}^{\circ }={105}^{\circ }(\text{mm})\text{atau}\\ & x={60}^{\circ }-{45}^{\circ }={15}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={60}^{\circ }\pm {45}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{{15}^{\circ },{105}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \cos x-\sqrt{3}\sin x=1\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & -\sqrt{3}\sin x+\cos x=1(\mathbf{\text{ingat}}:a=-\sqrt{3},b=1)\\ & -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \{\begin{array}{ll}k& =\sqrt{{(-\sqrt{3})}^2+{\left(1\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta ={300}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran IV, karena}a<0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \iff 2\cos (x-{300}^{\circ })=1\\ & \iff \cos (x-{300}^{\circ })={\displaystyle \frac{1}{2}}\\ & \iff \cos (x-{300}^{\circ })=\cos {60}^{\circ }\\ & \iff x-{300}^{\circ }=\pm {60}^{\circ }+k{.360}^{\circ }\\ & \iff x={300}^{\circ }\pm {60}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={300}^{\circ }+{60}^{\circ }={360}^{\circ }=0^{\circ }(\text{mm})\text{atau}\\ & x={300}^{\circ }-{60}^{\circ }={240}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={300}^{\circ }\pm {60}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{240}^{\circ },{360}^{\circ }\}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiasyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Noormandiri, B. K. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
3. 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.
4. Sukino. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.