$\begin{array}{ll}\\ 11.&\textrm{Himpunan penyelesaian persamaan}\\ &3\cos 2x+5\sin x+1=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}3\cos 2x+5\sin x+1&=0\\ 3\left ( 1-2\sin ^{2}x \right )+5\sin x+1&=0\\ -6\sin ^{2}+5\sin x+4&=0\\ 6\sin ^{2}x-5\sin x-4&=0\\ \left ( 3\sin x-4 \right )\left ( 2\sin x+1 \right )&=0\\ \sin x=\displaystyle \frac{4}{3}\: \: \textrm{atau}\: \: \sin x&=-\frac{1}{2}\\ \sin x&=\sin 150^{\circ}=\frac{5}{6}\pi \\ x&=\begin{cases} \displaystyle \frac{7}{6}\pi &+k.2\pi \\ \pi -\displaystyle \frac{7}{6}\pi & +k.2\pi \end{cases}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=\displaystyle \frac{7}{6}\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi \\ \textrm{saat}\: \: k&=1\\ x_{1}&=\color{red}\displaystyle \frac{7}{6}\pi +2\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi +2\pi =\frac{11}{6}\pi \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&\textrm{Himpunan penyelesaian dari}\\ &\sqrt{3}\sin 2x+2\cos ^{2}x=-1\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 240^{\circ},300^{\circ} \right \}\\ \textrm{b}.&\left \{ 30^{\circ},60^{\circ} \right \}\\ \textrm{c}.&\left \{ 150^{\circ},315^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 120^{\circ},300^{\circ} \right \}\\ \textrm{e}.&\left \{ 60^{\circ},150^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sqrt{3}\sin 2x+2\cos ^{2}x&=-1\\ \sqrt{3}\sin 2x+1+\cos 2x&=-1\\ \sqrt{3}\sin 2x+\cos 2x&=-2\\ \sqrt{\sqrt{3}^{2}+1^{2}}\cos \left ( 2x-\alpha \right )&=-2\\ a=x=1,\: \: b&=y=\sqrt{3}\\ \alpha &=\arctan \displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}\\ \alpha &=60^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( 2x-60^{\circ} \right )&=-2\\ \cos \left ( 2x-60^{\circ} \right )&=-1\\ \cos \left ( 2x-60^{\circ} \right )&=\cos 180^{\circ}\\ \left ( 2x-60^{\circ} \right )&=\pm 180^{\circ}+k.360^{\circ}\\ 2x&=60^{\circ}\pm 180^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 90^{\circ}+k.180^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=120^{\circ}\\ x_{2}&=-60^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=120^{\circ}+360^{\circ}=....\\ x_{2}&=-60^{\circ}+360^{\circ}=300^{\circ} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: 3\sin \theta +4\cos \theta =5 \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \sin \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0,3\\ \color{red}\textrm{b}.&0,60\\ \textrm{c}.&0,75\\ \textrm{d}.&0,80\\ \textrm{e}.&1,20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}3\sin \theta +4\cos \theta &=5\\ k\cos \left ( \theta -\alpha \right )&=5\\ a=x=4,\: &b=y=3\\ \theta &=\arctan \displaystyle \frac{b}{a}\\ \theta &=\arctan \displaystyle \frac{3}{4}\\ \color{black}\textrm{atau}\: \: &\\ \tan \theta &=\displaystyle \frac{3}{4}\\ \color{black}\textrm{maka}\: \: &\\ \sin \theta &=\displaystyle \frac{3}{\sqrt{3^{2}+4^{2}}}\\ &=\displaystyle \frac{3}{5}\\ &=\color{magenta}0,6 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: \tan \theta +\sec \theta =x \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \tan \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2x}{x^{2}-1}\\ \textrm{b}.&\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{c}.&\displaystyle \frac{x^{2}+1}{2x}\\ \color{red}\textrm{d}.&\displaystyle \frac{x^{2}-1}{2x}\\ \textrm{e}.&\displaystyle \frac{x^{2}-1}{x^{2}+1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\color{black}\textrm{Langkah}&\: \: 1\\ \tan \theta +\sec \theta &=x\\ \displaystyle \frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta }&=x\\ \displaystyle \frac{\sin \theta +1}{\cos \theta }&=x\\ \sin \theta +1&=x\cos \theta ...........1\\ \color{black}\textrm{Langkah}&\: \: 2\\ \left (\tan \theta +\sec \theta \right )^{2}&=x^{2}\\ \tan ^{2}\theta +2\tan \theta \sec \theta +\sec ^{2}\theta &=x^{2}\\ \sec ^{2}\theta -1+2\tan \theta \sec \theta &+\sec ^{2}\theta =x^{2}\\ 2\sec ^{2}\theta +2\tan \theta \sec \theta &=x^{2}+1\\ \displaystyle \frac{2}{\cos ^{2}\theta }+\frac{2\sin \theta }{\cos ^{2}\theta }&=x^{2}+1\\ 1+\sin \theta &=\left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}\theta.....2\\ \color{black}\textrm{Langkah}&\: \: 3\\ \left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}&=x\cos \theta \\ \cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{maka}\: \: &(\color{magenta}\textbf{dengan sisi segitiga})\\ \tan \theta =\displaystyle \frac{\sqrt{\left ( x^{2}+1 \right )^{2}-(2x)^{2}}}{2x}&=\displaystyle \frac{\sqrt{x^{4}+2x^{2}+1-4x^{2}}}{2x}\\ \tan \theta &=\displaystyle \frac{\sqrt{x^{4}-2x^{2}+1}}{2x}\\ &=\displaystyle \frac{\sqrt{\left ( x^{2}-1 \right )^{2}}}{2x}\\ &=\displaystyle \frac{x^{2}-1}{2x} \end{aligned} \end{array}$
$\begin{aligned}.\: \: \qquad \textbf{Sehingga}&\: \textbf{dari kasus di atas didapatkan}\\ \color{magenta}\sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1}\\ \color{magenta}\cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \color{magenta}\tan \theta &=\displaystyle \frac{x^{2}-1}{2x} \end{aligned}$
$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: \sec x +\tan x =\displaystyle \frac{3}{2} \: \: \textrm{untuk}\\ &0\leq x\leq \displaystyle \frac{\pi }{2}\: \: \textrm{maka nilai} \: \: \sin x=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{13}\\ \textrm{b}.&\displaystyle \frac{12}{13}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle \frac{2}{13}\\ \textrm{e}.&\displaystyle \frac{5}{12} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\sec x +\tan x &=\displaystyle \frac{3}{2}\\ \textrm{ingat saat}&\: \color{black}\textrm{mengerjakan soal no}.14\\ \sec \theta +\tan \theta &=x\\ \sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1},\: \: \color{black}\textrm{maka}\\ \sin x&=\color{magenta}\displaystyle \frac{\left ( \displaystyle \frac{3}{2} \right )^{2}-1}{\left (\displaystyle \frac{3}{2} \right )^{2}+1}\\ &=\displaystyle \frac{\displaystyle \frac{9}{4}-1}{\displaystyle \frac{9}{4}+1}\\ &=\frac{\displaystyle \frac{5}{4}}{\displaystyle \frac{13}{4}}\\ &=\color{magenta}\displaystyle \frac{5}{13} \end{aligned} \end{array}$
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