$\begin{array}{ll}\\ 11.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\cos x \\\\ &\textrm{b}.\quad \displaystyle y=\cos 2x \\\\ &\textrm{c}.\quad \displaystyle y=\cos \displaystyle \frac{1}{2}x \\\\ &\textrm{d}.\quad \displaystyle \color{red}y=2\cos 2x \\\\ &\textrm{e}.\quad \displaystyle y=2\cos \displaystyle \frac{1}{2}x \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\\ & \textbf{atas dan ke bawah}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ}=\pi ,\\ & \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\cos 2x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 12.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left (2x+\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{c}.\quad \displaystyle y=2\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=\sin \left (2x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin \displaystyle \left (x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{1}=360^{\circ}=2\pi ,\\ &\textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( x+\displaystyle kx \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: 90^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin (x+\displaystyle \frac{1}{2}\pi )^{\circ} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 13.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.
$\begin{array}{ll}\\ 14.&\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 \} \\\\ &\color{blue}\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}$.
$\begin{array}{ll}\\ 15.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos 2x-2\cos x=-1\: \: \textrm{untuk}\: \: 0< x< 2\pi \\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{3}{2}\pi, 2\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi, 2\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi ,\displaystyle \frac{3}{2}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos 2x-2\cos x=-1\\ &\Leftrightarrow \cos 2x-2\cos x+1=0\\ &\Leftrightarrow \left ( 2\cos ^{2}x-1 \right )-2\cos x+1=0 \\ &\Leftrightarrow 2\cos x\left ( \cos x-1 \right )=0\\ &\Leftrightarrow \cos x=0\: \: \textrm{atau}\: \: \cos x=1\\ &\Leftrightarrow \cos x=\cos \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: \cos x=\cos 0\\ &\Leftrightarrow x_{1,2}=\pm \displaystyle \frac{1}{2}\pi +k.2\pi \: \: \textrm{atau}\: \: x_{3}=k.2\pi\\ &\textrm{maka}\\ &k=0\Rightarrow x_{1}=-\displaystyle \frac{1}{2}\pi\: \: (\color{red}\textrm{tm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{1}{2}\pi\: \: (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{3}=\displaystyle 0\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1}=\displaystyle \frac{3}{2}\pi \quad (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{2}=\displaystyle \frac{5}{2}\pi\: \: (\color{red}\textrm{tm})\\ &\qquad\qquad x_{3}=\displaystyle 2\pi\: \: (\color{blue}\textrm{mm})\\ \end{aligned} \end{array} \end{array}$
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