$\begin{array}{ll}\\ 11.&\textrm{Periode dari fungsi}\: \: f(x)=-2\cos 3x\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad 90^{\circ} \\\\ &\textrm{b}.\quad 100^{\circ} \\\\ &\textrm{c}.\quad \color{red}120^{\circ} \\\\ &\textrm{d}.\quad 150^{\circ} \\\\ &\textrm{e}.\quad 180^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Periode dari}\: :\: f(x)=-2\cos 3x\\ &\textrm{adalah}\\ &=\displaystyle \frac{360^{\circ}}{3}\\ &=\color{red}120^{\circ}\\ &\\ &\textrm{Ingat bahwa}\\ &f(x)=a\cos bx,\: \: \textrm{maka periodenya}\\ &=\displaystyle \frac{360^{\circ}}{b} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 12.&\textrm{Perhatikanlah grafik berikut} \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{Gambar di atas adalah grafik fungsi dari}\\ &\textrm{a}.\quad f(x)=\cos 2x \\\\ &\textrm{b}.\quad f(x)=\cos 3x \\\\ &\textrm{c}.\quad \color{red}f(x)=3\cos x \\\\ &\textrm{d}.\quad f(x)=3\cos 3x \\\\ &\textrm{e}.\quad f(x)=\displaystyle \frac{1}{3}\cos x \\\\ &\textbf{Jawab}:\\ &\textrm{Gambar cukup jelas}\\ &\textrm{dengan periode}\: \: 360^{\circ}\\ &\textrm{gambar dari grafik}\: \: f(x)=\color{red}3\cos x \end{array}$.
$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }=\: ....\\\\ &\textrm{a}.\quad 4 \\\\ &\textrm{b}.\quad 2-\sqrt{2} \\\\ &\textrm{c}.\quad \sqrt{2}-2 \\\\ &\textrm{d}.\quad -4 \\\\ &\textrm{e}.\quad \color{red}-1 \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }\\ &=\displaystyle \frac{\sin 150^{\circ} +\tan 180^{\circ} +\cos 180^{\circ} }{\sin \displaystyle 90^{\circ} +\cos 360^{\circ} -3\cos \displaystyle 60^{\circ} }\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+0+(-1)}{1+1-3\left (\displaystyle \frac{1}{2} \right ) }\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}}{2-\displaystyle \frac{3}{2}}\\ &=-\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}-1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left ( x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=2\sin \left ( \displaystyle \frac{1}{2}\pi -x \right ) \\\\ &\textrm{c}.\quad \displaystyle \color{red}y=2\sin \left ( 2x+\displaystyle \frac{1}{6}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=-2\sin \left ( \displaystyle \frac{1}{2}\pi +x \right ) \\\\ &\textrm{e}.\quad \displaystyle y=-2\sin \left ( \displaystyle \frac{1}{2}\pi -2x \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi sinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: 2\pi,\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( 2x+k \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kirinya} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=-2\sin (3x+45)^{\circ} \\\\ &\textrm{b}.\quad \displaystyle y=-2\sin (3x-15)^{\circ} \\\\ &\textrm{c}.\quad \displaystyle y=-2\sin (3x-45)^{\circ} \\\\ &\textrm{d}.\quad \displaystyle y=2\sin (3x+15)^{\circ} \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin (3x-45)^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi sinus di geser ke}\: \: \textbf{kanan}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{3}=120^{\circ},\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin 3\left ( x-k \right )\\ &\textrm{dengan}\: \: -k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kanan}\: \: 15^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin 3(x-15)^{\circ}=2\sin (3x-45)^{\circ} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 16.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=-\cos (2x-30)^{\circ} \\\\ &\textrm{b}.\quad \displaystyle y=\sin (2x-60)^{\circ} \\\\ &\textrm{c}.\quad \displaystyle \color{red}y=\cos (2x+30)^{\circ} \\\\ &\textrm{d}.\quad \displaystyle y=\sin (2x-80)^{\circ} \\\\ &\textrm{e}.\quad \displaystyle y=\sin (2x+60)^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 1\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ},\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}\cos 2\left ( x+k \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: 15^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}\cos 2(x+15)^{\circ}=\sin (2x+30)^{\circ} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{1}{3}\pi ,\pi ,\displaystyle \frac{5}{3}\pi ,\displaystyle \frac{7}{3}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{2}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \\ &\Leftrightarrow \: \: 2x-\displaystyle \frac{1}{4}\pi=\displaystyle \frac{1}{4}\pi+k.\pi\\ &\Leftrightarrow \quad 2x =\displaystyle \frac{2}{4}\pi +k.\pi \\ &\Leftrightarrow \quad x =\displaystyle \frac{1}{4}\pi +k.\frac{\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{1}{4}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{\pi}{2}=\displaystyle \frac{3}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=2\Rightarrow x=\displaystyle \frac{1}{4}\pi+\pi =\displaystyle \frac{5}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=3\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{3\pi}{2}=\displaystyle \frac{7}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=4\Rightarrow x=\displaystyle \frac{1}{4}\pi+2\pi =\displaystyle \frac{9}{4}\pi \: \: (\color{red}\textrm{tidak memenuhi}) \\ \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle \frac{1}{4}\pi,\: \displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi \right \} \end{array} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{2} \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ 75^{\circ},285^{\circ} \right \}\\\\ &\textrm{b}.\quad \displaystyle \left \{ 75^{\circ},343^{\circ} \right \}\\\\ &\textrm{c}.\quad \displaystyle \left \{ 75^{\circ},344^{\circ} \right \}\\\\ &\textrm{d}.\quad \displaystyle \color{red}\left \{ 75^{\circ},345^{\circ} \right \}\\\\ &\textrm{e}.\quad \displaystyle \left \{ 75^{\circ},346^{\circ} \right \}\\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{2} \\ &\cos (x-30)^{\circ}= \cos 45^{\circ}\\ &\Leftrightarrow \: \: x-30^{\circ}=\pm 45^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x =30^{\circ}\pm 45^{\circ}+k.360^{\circ} \\ &k=0\Rightarrow x_{1}=75^{\circ}\: \: (\color{blue}\textrm{mm})\\ &\qquad\textrm{atau}\: \: x_{2}=-15^{\circ}\: \: \color{red}(tm)\\ &k=1\Rightarrow x_{1}=75^{\circ}+360^{\circ} \: \: (\color{red}\textrm{tm})\\ &\qquad \textrm{atau}\: \: \: x_{2}=-15^{\circ}+360^{\circ}=345^{\circ}\: \: \color{blue}(mm) \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle 75^{\circ},345^{\circ} \right \} \end{array} \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{3} \: \: \textrm{untuk}\: \: 0^{\circ}< x< 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ 100^{\circ},330^{\circ} \right \}\\\\ &\textrm{b}.\quad \displaystyle \color{red}\left \{ 30^{\circ},330^{\circ} \right \}\\\\ &\textrm{c}.\quad \displaystyle \left \{ 120^{\circ},300^{\circ} \right \}\\\\ &\textrm{d}.\quad \displaystyle \left \{ 60^{\circ},120^{\circ} \right \}\\\\ &\textrm{e}.\quad \displaystyle \left \{ 50^{\circ},300^{\circ} \right \}\\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos x^{\circ}= \displaystyle \frac{1}{2}\sqrt{3} \\ &\cos x^{\circ}= \cos 30^{\circ}\\ &\Leftrightarrow \quad x =\pm 30^{\circ}+k.360^{\circ} \\ &k=0\Rightarrow x_{1}=30^{\circ}\: \: (\color{blue}\textrm{mm})\\ &\qquad\textrm{atau}\: \: x_{2}=-30^{\circ}\: \: \color{red}(tm)\\ &k=1\Rightarrow x_{1}=30^{\circ}+360^{\circ}=390^{\circ} \: \: (\color{red}\textrm{tm})\\ &\qquad \textrm{atau}\: \: \: x_{2}=-30^{\circ}+360^{\circ}=330^{\circ}\: \: \color{blue}(mm) \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle 30^{\circ},330^{\circ} \right \} \end{array} \end{array}$
