Contoh Soal 2 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 6.&\textrm{Bentuk}\: \: \sqrt{3}\cos x-\sin x\: \: \textrm{untuk}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{dapat dinyatakan sebagai}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\cos \left ( x+\displaystyle \frac{\pi }{6} \right )\\ \textrm{b}.&2\cos \left ( x+\displaystyle \frac{7\pi }{6} \right ) \\ \color{red}\textrm{c}.&2\cos \left ( x-\displaystyle \frac{11\pi }{6} \right )\\ \textrm{d}.&2\cos \left ( x-\displaystyle \frac{7\pi }{6} \right )\\ \textrm{e}.&2\cos \left ( x-\displaystyle \frac{\pi }{6} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{c}\\ &\color{blue}\begin{aligned}\sqrt{3}\cos x-\sin x&=k\cos (x-\alpha )\\ (1)\qquad & (a,b)=\begin{cases} a &=\sqrt{3} \\ b &=-1 \end{cases}\\ \textrm{maka}\: &\textrm{titik ada dikadran IV}\\ (2)\: \quad k&=\sqrt{a^{2}+b^{2}}\\ &=\sqrt{\sqrt{3}^{2}+(-1)^{2}}\\ &=\sqrt{4}=2\\ (3)\quad \alpha &=\arctan \frac{b}{a}=\arctan \left (\frac{-1}{\sqrt{3}} \right )=-30^{\circ}\\ &=\left ( 360^{\circ}-30^{\circ} \right )=330^{\circ}=\displaystyle \frac{11}{6}\pi \\ \textrm{sehingga}&\\ \sqrt{3}\cos x-\sin x&=2\cos \left ( x-\displaystyle \frac{11}{6}\pi \right )\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai-nilai}\: \: x\: \: \textrm{yang terletak pada}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{yang memenuhi persamaan}\: \: \sqrt{3}\cos x+\sin x=\sqrt{2}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \color{red}\textrm{b}.&75^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{c}.&15^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \textrm{d}.&15^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{e}.&15^{\circ}\: \: \textrm{atau}\: \: 75^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\color{blue}\begin{aligned}\sqrt{3}\cos x+\sin x&=\sqrt{2}\\ \sqrt{\sqrt{3}^{2}+1^{2}}&\left ( \cos \left ( \alpha -\arctan \displaystyle \frac{1}{\sqrt{3}} \right ) \right )=\sqrt{2}\\ 2\cos \left ( x-30^{\circ} \right )&=\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\cos 45^{\circ}\\ \left ( x-30^{\circ} \right )&=\pm 45^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=75^{\circ}\\ x_{2}&=\color{red}-15^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=75^{\circ}+360^{\circ}=\color{red}435^{\circ}\\ x_{4}&=-15^{\circ}+360^{\circ}=345^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Diketahui fungsi trigonometri}\: \: f(x)=\displaystyle \frac{1}{2}\sin 3x\\ &\textrm{perhatikanlah pernyataan-pernyataan berikut}\\ &(1)\quad \textrm{hasil dari}\: \: f(0)+f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &(2)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )+f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(3)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )-f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(4)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{3} \right )-f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &\textrm{Pernyataan yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(1)\: \: \textrm{dan}\: \: (2)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \textrm{c}.&(1)\: \: \textrm{dan}\: \: (4)\\ \color{red}\textrm{d}.&(2)\: \: \textrm{dan}\: \: (3)\\ \textrm{e}.&(3)\: \: \textrm{dan}\: \: (4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{d}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\\ f(x)&=\displaystyle \frac{1}{2}\sin 3x\\ \textrm{maka}\qquad&\\ f(0)&=\displaystyle \frac{1}{2}\sin 3(0^{\circ})=0\\ f\left ( \displaystyle \frac{\pi }{3} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{3} \right )=0\\ f\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{6} \right )=\displaystyle \frac{1}{2}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Himpunan penyelesaian dari}\: \: \sin x-\sqrt{3}\cos x=-1\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0^{\circ},120^{\circ} \right \}\\ \textrm{b}.&\left \{ 90^{\circ},330^{\circ} \right \}\\ \textrm{c}.&\left \{ 60^{\circ},180^{\circ} \right \}\\ \textrm{d}.&\left \{ 90^{\circ},120^{\circ} \right \}\\ \color{red}\textrm{e}.&\left \{ 30^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\sin x-\sqrt{3}\cos x&=-1\\ \sqrt{1^{2}+\left ( -\sqrt{3} \right )^{2}}\cos \left ( x-\alpha \right )&=-1\\ a&=x=-\sqrt{3},\: \: b=y=1\\ &\textbf{kuadran II}\\ \alpha &=\arctan \left ( \displaystyle \frac{1}{-\sqrt{3}} \right )=-30^{\circ}\\ &=180^{\circ}-30^{\circ}=150^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( x-150^{\circ} \right )&=-1\\ \cos \left ( x-150^{\circ} \right )&=\displaystyle \frac{-1}{2}\\ \cos \left ( x-150^{\circ} \right )&=\cos 120^{\circ}\\ \left ( x-150^{\circ} \right )&=\pm 120^{\circ}+k.360^{\circ}\\ x&=150^{\circ}\pm 120^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: &k=0\\ x_{1}&=270^{\circ}\\ x_{2}&=30^{\circ}\\ \textrm{saat}\: \: &k=1\\ x_{3}&=270^{\circ}+360^{\circ}=....\\ x_{4}&=30^{\circ}+360^{\circ}=.... \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Nilai}\: \: \tan x\: \: \textrm{yang memenuhi persamaan}\\ &\cos 2x+7\cos x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{5}\sqrt{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\cos 2x+7\cos x-3&=0\\ 2\cos ^{2}x-1+7\cos x-3&=0\\ 2\cos ^{2}x+7\cos x-4&=0\\ \left ( \cos x+4 \right )\left (2 \cos -1 \right )&=0\\ \color{red}\cos x=-4\: \: \color{black}\textrm{atau}\: \: \color{magenta}\cos x&=\displaystyle \frac{1}{2}=\cos 60^{\circ}\\ x&=60^{\circ},\\ \textrm{maka}&\\ \tan 60^{\circ}&=\sqrt{3}\\\\ \textrm{ dan ingat bahwa}&\: \: \cos x=-4\: \: \color{red}\textrm{tidak memenuhi} \end{aligned} \end{array}$