Latihan Soal 3 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 21.&\textrm{Jika diketahui}\: \: \tan \left ( 3x+60^{0} \right )=\sqrt{3},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\textrm{tersebut adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&90^{0}\\ \textrm{b}.&110^{0}\\ \textrm{c}.&\color{red}120^{0}\\ \textrm{d}.&130^{0}\\ \textrm{e}.&230^{0} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\tan \left ( 3x+60^{0} \right )&=\sqrt{3}\\ \tan \left ( 3x+60^{0} \right )&=\tan 60^{0}\\ \left ( 3x+60^{0} \right )&=60^{0}+k.180^{0}\\ 3x&=k.180^{0}\\ x&=k.60^{0}\\ k=0\rightarrow x&=0^{0}\\ k=1\rightarrow x&=60^{0}\\ k=2\rightarrow x&=120^{0}\\ k=3\rightarrow x&=180^{0}\\ k=4\rightarrow x&=240^{0}\\ k=5\rightarrow x&=300^{0}\\ k=\cdots \rightarrow x&=\cdots \\ \cdots \cdots \rightarrow &=....\textbf{dst} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Himpunan penyelesaian dari }\\ & \sqrt{3}\tan x=-1,\quad 0^{0}\leq x\leq 360^{0}\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 30^{0},150^{0} \right \}\\ \textrm{b}.&\left \{ 150^{0},210^{0} \right \}\\ \textrm{c}.&\color{red}\left \{ 150^{0},330^{0} \right \}\\ \textrm{d}.&\left \{ 120^{0},300^{0} \right \}\\ \textrm{e}.&\left \{ 60^{0},240^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{3}\tan x&=-1\\ \tan x&=-\displaystyle \frac{1}{\sqrt{3}}=-\frac{1}{3}\sqrt{3}\\ &\qquad \color{blue}\textrm{nilai di kuadran II}\\ \tan x&=\tan 150^{0}\\ x&=150^{0}+k.180^{0}\\ k=0,\Rightarrow x&=150^{0}\\ k=1,\Rightarrow x&=330^{0}\\ k=2,\Rightarrow x&=510^{0},\quad \textrm{tidak}\: \textrm{memenuhi}\: \textrm{(tm)}\\ k=3,\quad \textbf{dst}&=\textrm{(tm)}\\ \therefore \textbf{HP}&=\color{red}\left \{ 150^{0},330^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Himpunan penyelesaian dari }\\ & \tan 2x^{0}=-\displaystyle \frac{1}{3}\sqrt{3},\quad 0^{0}\leq x\leq 180^{0}\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ 75^{0},165^{0} \right \}\\ \textrm{b}.&\left \{ 60^{0},150^{0} \right \}\\ \textrm{d}.&\left \{ 30^{0},120^{0} \right \}\\ \textrm{c}.&\left \{ 45^{0},135^{0} \right \}\\ \textrm{e}.&\left \{ 15^{0},105^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{a}\\ &\begin{aligned}\tan 2x^{0}&=-\displaystyle \frac{1}{3}\sqrt{3}\\ \tan 2x^{0}&=\tan \left ( 180^{0}-30^{0} \right )\\ &\textrm{nilai tan negatif paling kecil }\\ &\textrm{berada di kuadran II}\\ 2x^{0}&=150^{0}+k.180^{0}\\ x^{0}&=75^{0}+k.90^{0}\\ k=0,\Rightarrow x^{0}&=75^{0}\\ k=1,\Rightarrow x^{0}&=75^{0}+90^{0}=165^{0}\\ k=2,\Rightarrow x^{0}&=255^{0}\\ &\textrm{tidak memenuhi, karena }\\ &\textrm{berada di luar batas interval}\\ \therefore \textbf{HP}&=\color{red}\left \{ 75^{0},165^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Himpunan penyelesaian dari }\\ & \cos \left ( x-30^{0} \right )=-\sin 50^{0},\quad 0^{0}\leq x\leq 360^{0}\\ & \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 90^{0} \right \}\\ \textrm{b}.&\left \{ 140^{0},150^{0} \right \}\\ \textrm{c}.&\color{red}\left \{ 170^{0},250^{0} \right \}\\ \textrm{d}.&\left \{ 80^{0},280^{0} \right \}\\ \textrm{e}.&\left \{ 20^{0},340^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\cos \left ( x-30^{0} \right )&=-\sin 50^{0}\\ \cos \left ( x-30^{0} \right )&=\cos \left ( 90^{0}+50^{0} \right )\\ \cos \left ( x-30^{0} \right )&=\cos 140^{0}\\ x-30^{0}&=\pm 140^{0}+k.360^{0}\\ x&=30^{0}\pm 140^{0}+k.360^{0}\\ k=0,\Rightarrow x_{1}&=30^{0}+140^{0}=170^{0}\\ \Rightarrow x_{2}&=30^{0}-140^{0}=-110^{0}\: \: (\textrm{tm})\\ k=1,\Rightarrow x_{1}&=30^{0}+140^{0}+360^{0}=530^{0}\: \: (\textrm{tm})\\ \Rightarrow x_{2}&=30^{0}-140^{0}+360^{0}=250^{0}\\ \therefore \textbf{HP}&=\color{red}\left \{ 170^{0},250^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Himpunan penyelesaian dari}\\ &\sin 2x=\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 30^{\circ},210^{\circ} \right \}\\ \textrm{b}.&\left \{ 60^{\circ},240^{\circ} \right \}\\ \textrm{c}.&\left \{ 30^{\circ},60^{\circ},210^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ} \right \}\\ \textrm{e}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sin 2x&=\displaystyle \frac{1}{2}\sqrt{3}\\ \sin 2x&=\sin 60^{\circ}\\ 2x&=\begin{cases} 60^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-60^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ x&=\begin{cases} 30^{\circ} & +k.180^{\circ}\\ 60^{\circ} & +k.180^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=0\\ x&=\begin{cases} 30^{\circ} & \\ 60^{\circ} & \end{cases}\\ \textrm{saat}&\: \: k=1\\ x&=\begin{cases} 30^{\circ} & +1.180^{\circ}=210^{\circ}\\ 60^{\circ} & +1.180^{\circ}=240^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=2\\ x&=\begin{cases} 30^{\circ} & +2.180^{\circ}=\color{red}390^{\circ}\\ 60^{\circ} & +2.180^{\circ}=\color{red}420^{\circ} \end{cases}\\ \color{red}\textrm{kedua}&\color{red}\textrm{nnya tidak memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 26.&\textrm{Himpunan penyelesaian dari}\\ &\tan 2x-\sqrt{3}=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \color{red}\textrm{b}.&\left \{ 30^{\circ},120^{\circ},210^{\circ},300^{\circ} \right \}\\ \textrm{c}.&\left \{ 45^{\circ},135^{\circ},225^{\circ},315^{\circ} \right \}\\ \textrm{d}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \textrm{e}.&\left \{ 15^{\circ},30^{\circ},45^{\circ},60^{\circ},75^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\tan 2x&-\sqrt{3}=0\\ \tan 2x&=\sqrt{3}\\ \tan 2x&=\tan 60^{\circ}\\ 2x&=60+k.180^{\circ}\\ x&=30^{\circ}+k.90^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=30^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=30^{\circ}+90^{\circ}=120^{\circ}\\ \textrm{saat}&\: \: k=2\\ x&=30^{\circ}+180^{\circ}=210^{\circ}\\ \textrm{saat}&\: \: k=3\\ x&=30^{\circ}+270^{\circ}=300^{\circ}\\ \textrm{saat}&\: \: k=4\\ x&=30^{\circ}+360^{\circ}=\color{red}390^{\circ}\\ \color{red}\textrm{tidak}&\: \color{red}\textrm{memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Himpunan penyelesaian dari}\\ &\cos 3x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 180^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 40^{\circ},80^{\circ} \right \}\\ \textrm{b}.&\left \{ 50^{\circ},70^{\circ} \right \}\\ \textrm{c}.&\left \{ 40^{\circ},70^{\circ},80^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 50^{\circ},70^{\circ},170^{\circ} \right \}\\ \textrm{e}.&\left \{ 50^{\circ},80^{\circ},170^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\cos 3x&=-\displaystyle \frac{1}{2}\sqrt{3}\\ \cos 3x&=-\cos 30^{\circ}\\ \cos 3x&=\cos \left (180^{\circ}-30^{\circ} \right )=\cos 150^{\circ}\\ 3x&=\pm 150^{\circ}+k.360^{\circ}\\ x&=\pm 50^{\circ}+k.120^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=\pm 50^{\circ}\: \rightarrow x=50^{\circ}\: \: (\textrm{mm})\\ \textrm{saat}&\: \: k=1\\ x&=\pm 50^{\circ}+120^{\circ}=\begin{cases} 170^{\circ} & (\textrm{mm}) \\ 70^{\circ} & (\textrm{mm}) \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Jika diketahui}\: \: \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \textrm{dengan}\: \: 0<\beta <\displaystyle \frac{\pi }{2},\\ &\textrm{maka himpunan harga}\: \: \sin \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &&&&\textbf{(SIMAK UI 2009)} \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\sin \beta -\tan \beta -2\cos \beta +2&=0\\ \sin \beta -\displaystyle \frac{\sin \beta }{\cos \beta } -2\cos \beta +2&=0\\ \sin \beta \cos \beta -\sin \beta -2\cos^{2} \beta +2\cos \beta &=0\\ \sin \beta \left ( \cos \beta -1 \right )-2\cos \beta \left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )\left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )=0\: \textbf{(mm)}&\\ \color{red}\textrm{atau}\: \: \color{black}\left ( \cos \beta -1 \right )&=0\: \textbf{(tm)}\\ \color{blue}\textrm{Selanjutnya adalah},\qquad\qquad\quad&\\ \left ( \sin \beta -2\cos \beta \right )&=0\\ \sin \beta =2\cos \beta&\\ \displaystyle \frac{\sin \beta }{\cos \beta }&=2\\ \tan \beta &=2=\displaystyle \frac{2}{1},\\ \textrm{buatlah ilustrasi}& \\ \textrm{dengan membuat segitiga siku-siku}&\\ \textrm{Sehingga akan didapatkan nilai}&\\ \sin \beta =\color{red}\displaystyle \frac{2}{5}\sqrt{5}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Penyelesaian umum untuk nilai}\: \: \theta \\ & \textrm{yang memenuhi persamaan}\\ &\sin \theta =\displaystyle \frac{1}{2},\: \: \tan \theta =\displaystyle \frac{1}{\sqrt{3}} \: \: \textrm{adalah}\, ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}2n\pi +\displaystyle \frac{\pi }{6}\\\\ \textrm{b}.&2n\pi +\displaystyle \frac{\pi }{5}\\\\ \textrm{c}.&2n\pi +\displaystyle \frac{\pi }{4} \\\\ \textrm{d}.&2n\pi +\displaystyle \frac{\pi }{3}\\\\ \textrm{e}.&2n\pi +\displaystyle \frac{\pi }{2} \\ \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\, ;\\ &\begin{cases} \sin x & =\sin \theta \\ x& = \theta +k.2\pi \\ \tan x & =\tan \theta \\ x&=\theta +k.\pi \end{cases}\\ \textrm{Karena}\quad&\sin x=\sin \theta =\displaystyle \frac{1}{2}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{Demikian}&\: \textrm{juga untuk nilai}\: \: \tan \theta ,\: \: \textrm{yaitu}:\\ &\tan x=\tan \theta =\displaystyle \frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{maka dari}&\: \textrm{sini akan diperoleh}\\ &\textbf{penyelesaian umumnya}\\ \textrm{yaitu}\quad &:\color{red}2n\pi +\displaystyle \frac{\pi }{6} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Jika diketahui}\\ & f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )=\tan x,\: \: \pi \leq x\leq 2\pi ,\\ &\textrm{maka nilai}\: \: f(3)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}0\\ \textrm{b}.&1\\ \textrm{c}.&\displaystyle \frac{\pi }{2}\\ \textrm{d}.&\pi \\ \textrm{e}.&2\pi \\ \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\textrm{Diketahui bahwa}\: &\\ f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )&=\tan x,\: \: \pi \leq x\leq 2\pi\\ f(3)&=....\\ \textrm{maka selanjutnya}\, &\\ \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}}&=3\\ \sqrt{\sin ^{2}x+4}&=2\\ \sin ^{2}x+4&=4\\ \sin ^{2}x&=0\\ \sin x&=\begin{cases} x_{1}=0^{\circ} &+k.2\pi \\ x_{2}=180^{\circ} & +k.2\pi \end{cases}\\ \tan \pi =\tan 2\pi &=0\\ \therefore \quad f(3)&=\color{red}0 \end{aligned} \end{array}$