Contoh Soal 1 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 1.&\textrm{Himpunan penyelesaian dari}\\ &\sin 2x=\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 30^{\circ},210^{\circ} \right \}\\ \textrm{b}.&\left \{ 60^{\circ},240^{\circ} \right \}\\ \textrm{c}.&\left \{ 30^{\circ},60^{\circ},210^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ} \right \}\\ \textrm{e}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sin 2x&=\displaystyle \frac{1}{2}\sqrt{3}\\ \sin 2x&=\sin 60^{\circ}\\ 2x&=\begin{cases} 60^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-60^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ x&=\begin{cases} 30^{\circ} & +k.180^{\circ}\\ 60^{\circ} & +k.180^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=0\\ x&=\begin{cases} 30^{\circ} & \\ 60^{\circ} & \end{cases}\\ \textrm{saat}&\: \: k=1\\ x&=\begin{cases} 30^{\circ} & +1.180^{\circ}=210^{\circ}\\ 60^{\circ} & +1.180^{\circ}=240^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=2\\ x&=\begin{cases} 30^{\circ} & +2.180^{\circ}=\color{red}390^{\circ}\\ 60^{\circ} & +2.180^{\circ}=\color{red}420^{\circ} \end{cases}\\ \color{red}\textrm{kedua}&\color{red}\textrm{nnya tidak memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Himpunan penyelesaian dari}\\ &\tan 2x-\sqrt{3}=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \color{red}\textrm{b}.&\left \{ 30^{\circ},120^{\circ},210^{\circ},300^{\circ} \right \}\\ \textrm{c}.&\left \{ 45^{\circ},135^{\circ},225^{\circ},315^{\circ} \right \}\\ \textrm{d}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \textrm{e}.&\left \{ 15^{\circ},30^{\circ},45^{\circ},60^{\circ},75^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\tan 2x&-\sqrt{3}=0\\ \tan 2x&=\sqrt{3}\\ \tan 2x&=\tan 60^{\circ}\\ 2x&=60+k.180^{\circ}\\ x&=30^{\circ}+k.90^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=30^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=30^{\circ}+90^{\circ}=120^{\circ}\\ \textrm{saat}&\: \: k=2\\ x&=30^{\circ}+180^{\circ}=210^{\circ}\\ \textrm{saat}&\: \: k=3\\ x&=30^{\circ}+270^{\circ}=300^{\circ}\\ \textrm{saat}&\: \: k=4\\ x&=30^{\circ}+360^{\circ}=\color{red}390^{\circ}\\ \color{red}\textrm{tidak}&\: \color{red}\textrm{memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Himpunan penyelesaian dari}\\ &\cos 3x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 180^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 40^{\circ},80^{\circ} \right \}\\ \textrm{b}.&\left \{ 50^{\circ},70^{\circ} \right \}\\ \textrm{c}.&\left \{ 40^{\circ},70^{\circ},80^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 50^{\circ},70^{\circ},170^{\circ} \right \}\\ \textrm{e}.&\left \{ 50^{\circ},80^{\circ},170^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\cos 3x&=-\displaystyle \frac{1}{2}\sqrt{3}\\ \cos 3x&=-\cos 30^{\circ}\\ \cos 3x&=\cos \left (180^{\circ}-30^{\circ} \right )=\cos 150^{\circ}\\ 3x&=\pm 150^{\circ}+k.360^{\circ}\\ x&=\pm 50^{\circ}+k.120^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=\pm 50^{\circ}\: \rightarrow x=50^{\circ}\: \: (\textrm{mm})\\ \textrm{saat}&\: \: k=1\\ x&=\pm 50^{\circ}+120^{\circ}=\begin{cases} 170^{\circ} & (\textrm{mm}) \\ 70^{\circ} & (\textrm{mm}) \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2\cos ^{2}x+\cos x-1=0\: \: \textrm{untuk}\: \: 0\leq x\leq \pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \pi \\ \textrm{b}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \\ \textrm{c}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{d}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{e}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}2\cos ^{2}x+\cos x-1&=0\\ \left (2\cos x-1 \right )\left (\cos x+1 \right )&=0\\ \cos x=\displaystyle \frac{1}{2}\: \: \color{magenta}\textrm{atau}\: \: &\cos x=-1\\ \cos x=\cos 60^{\circ}=\cos \frac{1}{3}\pi \: \: &\\ \color{magenta}\textrm{atau}\: \: \cos x&=\cos 180^{\circ}=\cos \pi \\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan ^{2}x-\tan x-6=0\: \: \textrm{pada}\: \: 0\leq x\leq \pi ,\\ &\textrm{maka himpunan nilai}\: \: \sin x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5} \right \} \\ \textrm{c}.&\left \{ -\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \\ \textrm{d}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5} \right \} \\ \textrm{e}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\tan ^{2}x-\tan x-6&=0\\ \left (\tan x-3 \right )\left (\tan x+2 \right )&=0\\ \tan x=3\: \: \textrm{atau}\: \: &\tan x=-2\\ \tan x=\displaystyle \frac{3}{1}\: \: \textrm{atau}\: \: &\tan x=\frac{-2}{1}\\ \sin x=\displaystyle \frac{3}{\sqrt{1^{2}+3^{2}}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{1^{2}+2^{2}}}\\ \sin x=\displaystyle \frac{3}{\sqrt{10}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{5}}\\ \sin x=\displaystyle \frac{3}{10}\sqrt{10}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{5}\sqrt{5} \end{aligned} \end{array}$