$\begin{array}{ll}\\ 16.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\qquad\quad 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3} \\ & \textrm{untuk}\: \: 0\leq x\leq \pi \: \: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ \displaystyle \frac{1}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{5}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\frac{\sqrt{3}}{3}\\ &(\textrm{kuadran IV, karena Y negatif, X positif})\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\tan \displaystyle \frac{1}{6}\pi ,\: \: \textbf{menjadi}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=\tan \left ( 2\pi -\displaystyle \frac{1}{6}\pi \right )=\tan \displaystyle \frac{11}{6}\pi \\ &\left (2x-\displaystyle \frac{1}{3}\pi \right )=\displaystyle \frac{11}{6}\pi\\ &\Leftrightarrow \: \: 2x=\displaystyle \frac{1}{3}\pi+\displaystyle \frac{11}{6}\pi +k.\pi =\displaystyle \frac{13}{6}\pi +k.\pi \\ &\Leftrightarrow \: \: x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{k.\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{13}{12}\pi=\displaystyle \frac{1}{12}\pi \: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{1}{2}\pi =\displaystyle \frac{19}{12}\pi=\displaystyle \frac{7}{12}\pi \quad (\color{blue}\textrm{mm})\\ &k=2\Rightarrow x=\displaystyle \frac{13}{12}\pi +\pi \quad \color{red}\textrm{tidak memenuhi} \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{1}{12}\pi,\: \displaystyle \frac{7}{12}\pi \right \} \end{array} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \cos x+\sin x=\displaystyle \frac{1}{2}\sqrt{6}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{24}\pi &&&\textrm{d}.&\displaystyle \frac{1}{8}\pi \\\\ \textrm{b}.&\displaystyle \frac{1}{15}\pi&\textrm{c}.&\displaystyle \color{red}\frac{1}{12}\pi&\textrm{e}.&\displaystyle \frac{1}{6}\pi \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui bahwa}\\ &\sin x+\cos x=\displaystyle \frac{1}{2}\sqrt{6}\quad \left (\textbf{ingat}:a=1,\: b=1 \right )\\ &\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\begin{cases} k & =\sqrt{1^{2}+1^{2}}=\sqrt{2} \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{1}=1\Rightarrow \theta =45^{\circ}=\displaystyle \frac{1}{4}\pi \end{cases}\\ &\qquad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a,b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \sqrt{2}\cos\left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \: \: \, \cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{6}}{\sqrt{2}}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \: \: \,\cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\cos \displaystyle \frac{1}{6}\pi \\ &\Leftrightarrow \quad x-\displaystyle \frac{1}{4}\pi =\pm \displaystyle \frac{1}{6}\pi +k.2\pi \\ &\Leftrightarrow \quad x=\displaystyle \frac{1}{4}\pi\pm \displaystyle \frac{1}{6}\pi+k.2\pi \\ &\Leftrightarrow \quad x_{1}=\displaystyle \frac{5}{12}\pi+k.2\pi \: \: \textbf{atau}\\ &\: \: \: \quad\quad x_{2}=\displaystyle \frac{1}{12}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{5}{12}\pi\qquad (\color{blue}\textrm{memenuhi})\\ &\: \: \qquad\Rightarrow x_{2}=\displaystyle \color{red}\frac{1}{12}\pi\qquad \color{black}(\color{blue}\textrm{memenuhi})\\ &\textrm{Langkah berikutnya tidak diperlukan}\\ &\textrm{karena jawaban sudah kita dapatkan}\\ &\textrm{yaitu}:\: \: \color{red}\displaystyle \frac{1}{12}\pi \end{aligned} \end{array} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Himpunan penyelesaian persamaan}\\ &\qquad\: \: \cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1\\ &\textrm{untuk}\: \: 0\leq x< 360\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}\left \{ 0,240 \right \} &&&\textrm{d}.&\displaystyle \left \{ 180,240 \right \} \\\\ \textrm{b}.&\displaystyle \left \{ 150,270 \right \}&\textrm{c}.&\displaystyle \left \{ 180,300 \right \}&\textrm{e}.&\displaystyle \left \{ 210,270 \right \} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui dari soal bahwa}\\ &\cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1,\\ &\textrm{lalu kita ubah posisinya menjadi}\\\\ &-\sqrt{3}\sin x+\cos x=1\: \: \left (\textbf{ingat}:a=-\sqrt{3},\: b=1 \right )\\ &-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\begin{cases} k & =\sqrt{\left ( -\sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta =300^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran IV, karena}\: \: a<0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\Leftrightarrow 2\cos\left ( x-300^{\circ} \right )=1\\ &\Leftrightarrow \: \: \, \cos \left ( x-300^{\circ} \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-300^{\circ} \right )=\cos 60^{\circ}\\ &\Leftrightarrow \quad x-300^{\circ} =\pm 60^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=300^{\circ}\pm 60^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x_{1}=300^{\circ}+60^{\circ}=360^{\circ}=0^{\circ}\: \: (\color{blue}\textrm{mm})\\ & \qquad\qquad\quad\color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=300^{\circ}-60^{\circ}=240^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=300^{\circ}\pm 60^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{0^{\circ},240^{\circ} \right \} \end{array} \end{array}$
$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: \alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah sudut} \: \: \textbf{lancip}\\ &\textrm{Nilai dari}\: \: \cos \left ( \alpha +\beta \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{1}{2}&\textrm{e}.&\color{red}\displaystyle 0 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\bullet \quad\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut} \: \: \textbf{lancip}\\ &\qquad \textrm{akibatnya semua sudut dikuadran I}\\ &\qquad \textrm{sehingga}\color{purple}\begin{cases} \sin & =+ \\ \cos & =+ \\ \tan & =+ \end{cases}\\ &\textrm{ditanya}\: \: \cos \left ( \alpha +\beta \right ),\: \: \textrm{maka}\\ &\textrm{sebagai langkah awal kita adalah}:\\ &\cos \left ( \alpha -\beta \right )=\cos \left ( \displaystyle \frac{\pi }{3} \right )\\ &\Leftrightarrow \: \cos \alpha \cos \beta +\sin \alpha \sin \beta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta+\displaystyle \frac{1}{4} =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta =\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}=\displaystyle \frac{1}{4}\\ &\textbf{Selanjutnya nilai dari}\\ &\cos \left ( \alpha +\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}=0\\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \sin 75^{\circ}-\sin 165^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\sqrt{2}&&&\textrm{d}.&\color{red}\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{4}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{4}\sqrt{6}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sin 75^{\circ}-\sin 165^{\circ}\\ &=2\cos \left ( \displaystyle \frac{75^{\circ}+165^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-165^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{240^{\circ}}{2}\sin \left (-\displaystyle \frac{90^{\circ}}{2} \right ) \\ &=2\cos 120^{\circ}\sin \left ( -45^{\circ} \right )\\ &=2\left (-\cos 60^{\circ} \right )\left (- \sin 45^{\circ} \right )\\ &=2\left ( -\displaystyle \frac{1}{2} \right )\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.
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