Contoh Soal 1 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 1.& \text{Himpunan penyelesaian dari}\\ & \sin 2x={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^{\circ },{210}^{\circ }\}\\ \text{b}.& \{{60}^{\circ },{240}^{\circ }\}\\ \text{c}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ }\}\\ \text{d}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sin 2x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \sin 2x& =\sin {60}^{\circ }\\ 2x& =\{\begin{array}{ll}{60}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{60}^{\circ })& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.180}^{\circ }\\ {60}^{\circ }& +k{.180}^{\circ }\end{array}\\ \text{saat}& k=0\\ x& =\{\begin{array}{ll}{30}^{\circ }& \\ {60}^{\circ }& \end{array}\\ \text{saat}& k=1\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{1.180}^{\circ }={210}^{\circ }\\ {60}^{\circ }& +{1.180}^{\circ }={240}^{\circ }\end{array}\\ \text{saat}& k=2\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{2.180}^{\circ }={390}^{\circ }\\ {60}^{\circ }& +{2.180}^{\circ }={420}^{\circ }\end{array}\\ \text{kedua}& \text{nnya tidak memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Himpunan penyelesaian dari}\\ & \tan 2x-\sqrt{3}=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\\ \text{c}.& \{{45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{e}.& \{{15}^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ },{75}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\tan 2x& -\sqrt{3}=0\\ \tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& =60+k{.180}^{\circ }\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\\ \text{saat}& k=1\\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\\ \text{saat}& k=2\\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\\ \text{saat}& k=3\\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\\ \text{saat}& k=4\\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\\ \text{tidak}& \text{memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Himpunan penyelesaian dari}\\ & \cos 3x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {180}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{40}^{\circ },{80}^{\circ }\}\\ \text{b}.& \{{50}^{\circ },{70}^{\circ }\}\\ \text{c}.& \{{40}^{\circ },{70}^{\circ },{80}^{\circ }\}\\ \text{d}.& \{{50}^{\circ },{70}^{\circ },{170}^{\circ }\}\\ \text{e}.& \{{50}^{\circ },{80}^{\circ },{170}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos 3x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 3x& =-\cos {30}^{\circ }\\ \cos 3x& =\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 3x& =\pm {150}^{\circ }+k{.360}^{\circ }\\ x& =\pm {50}^{\circ }+k{.120}^{\circ }\\ \text{saat}& k=0\\ x& =\pm {50}^{\circ }\to x={50}^{\circ }(\text{mm})\\ \text{saat}& k=1\\ x& =\pm {50}^{\circ }+{120}^{\circ }=\{\begin{array}{ll}{170}^{\circ }& (\text{mm})\\ {70}^{\circ }& (\text{mm})\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & 2{\cos }^{2}x+\cos x-1=0\text{untuk}0\le x\le \pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}\pi \text{dan}\pi }\\ \text{b}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{2}{3}\pi }\\ \text{c}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{3}{4}\pi }\\ \text{d}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{3}{4}\pi }\\ \text{e}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{2}{3}\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}2{\cos }^{2}x+\cos x-1& =0\\ (2\cos x-1)(\cos x+1)& =0\\ \cos x={\displaystyle \frac{1}{2}\text{atau}}& \cos x=-1\\ \cos x=\cos {60}^{\circ }=\cos \frac{1}{3}\pi & \\ \text{atau}\cos x& =\cos {180}^{\circ }=\cos \pi \end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Untuk}x\text{yang memenuhi persamaan}\\ & {\tan }^{2}x-\tan x-6=0\text{pada}0\le x\le \pi ,\\ & \text{maka himpunan nilai}\sin x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\\ \text{b}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5}}\right\}\\ \text{c}.& \{-{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\}\\ \text{d}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5}}\right\}\\ \text{e}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\tan }^{2}x-\tan x-6& =0\\ (\tan x-3)(\tan x+2)& =0\\ \tan x=3\text{atau}& \tan x=-2\\ \tan x={\displaystyle \frac{3}{1}\text{atau}}& \tan x=\frac{-2}{1}\\ \sin x={\displaystyle \frac{3}{\sqrt{1^2+3^2}}\text{atau}}& \sin x=\frac{2}{\sqrt{1^2+2^2}}\\ \sin x={\displaystyle \frac{3}{\sqrt{10}}\text{atau}}& \sin x=\frac{2}{\sqrt{5}}\\ \sin x={\displaystyle \frac{3}{10}\sqrt{10}\text{atau}}& \sin x=\frac{2}{5}\sqrt{5}\end{array}\end{array}$