Contoh Soal 2 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

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

$\begin{array}{l}\\ 7.& \text{Nilai-nilai}x\text{yang terletak pada}0\le x\le 2\pi ,\\ & \text{yang memenuhi persamaan}\sqrt{3}\cos x+\sin x=\sqrt{2}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {75}^{\circ }\text{atau}{285}^{\circ }\\ \text{b}.& {75}^{\circ }\text{atau}{345}^{\circ }\\ \text{c}.& {15}^{\circ }\text{atau}{285}^{\circ }\\ \text{d}.& {15}^{\circ }\text{atau}{345}^{\circ }\\ \text{e}.& {15}^{\circ }\text{atau}{75}^{\circ }\end{array}\\ \\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}\sqrt{3}\cos x+\sin x& =\sqrt{2}\\ \sqrt{{\sqrt{3}}^2+1^2}& (\cos (\alpha -\arctan {\displaystyle \frac{1}{\sqrt{3}}}))=\sqrt{2}\\ 2\cos (x-{30}^{\circ })& =\sqrt{2}\\ \cos (x-{30}^{\circ })& ={\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}}\\ \cos (x-{30}^{\circ })& =\cos {45}^{\circ }\\ (x-{30}^{\circ })& =\pm {45}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ \text{saat}k& =0\\ x_1& ={75}^{\circ }\\ x_2& =-{15}^{\circ }\\ \text{saat}k& =1\\ x_3& ={75}^{\circ }+{360}^{\circ }={435}^{\circ }\\ x_4& =-{15}^{\circ }+{360}^{\circ }={345}^{\circ }\end{array}\end{array}$

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

$\begin{array}{l}\\ 9.& \text{Himpunan penyelesaian dari}\sin x-\sqrt{3}\cos x=-1\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{0^{\circ },{120}^{\circ }\}\\ \text{b}.& \{{90}^{\circ },{330}^{\circ }\}\\ \text{c}.& \{{60}^{\circ },{180}^{\circ }\}\\ \text{d}.& \{{90}^{\circ },{120}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\sin x-\sqrt{3}\cos x& =-1\\ \sqrt{1^2+{(-\sqrt{3})}^2}\cos (x-\alpha )& =-1\\ a& =x=-\sqrt{3},b=y=1\\ & \mathbf{\text{kuadran II}}\\ \alpha & =\arctan \left({\displaystyle \frac{1}{-\sqrt{3}}}\right)=-{30}^{\circ }\\ & ={180}^{\circ }-{30}^{\circ }={150}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (x-{150}^{\circ })& =-1\\ \cos (x-{150}^{\circ })& ={\displaystyle \frac{-1}{2}}\\ \cos (x-{150}^{\circ })& =\cos {120}^{\circ }\\ (x-{150}^{\circ })& =\pm {120}^{\circ }+k{.360}^{\circ }\\ x& ={150}^{\circ }\pm {120}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x_1& ={270}^{\circ }\\ x_2& ={30}^{\circ }\\ \text{saat}& k=1\\ x_3& ={270}^{\circ }+{360}^{\circ }=....\\ x_4& ={30}^{\circ }+{360}^{\circ }=....\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai}\tan x\text{yang memenuhi persamaan}\\ & \cos 2x+7\cos x-3=0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& {\displaystyle \frac{1}{3}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{5}\sqrt{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\cos 2x+7\cos x-3& =0\\ 2{\cos }^{2}x-1+7\cos x-3& =0\\ 2{\cos }^{2}x+7\cos x-4& =0\\ (\cos x+4)(2\cos -1)& =0\\ \cos x=-4\text{atau}\cos x& ={\displaystyle \frac{1}{2}=\cos {60}^{\circ }}\\ x& ={60}^{\circ },\\ \text{maka}& \\ \tan {60}^{\circ }& =\sqrt{3}\\ \\ \text{ dan ingat bahwa}& \cos x=-4\text{tidak memenuhi}\end{array}\end{array}$