Latihan Soal 4 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2\cos ^{2}x+\cos x-1=0\: \: \textrm{untuk}\: \: 0\leq x\leq \pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \pi \\ \textrm{b}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \\ \textrm{c}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{d}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{e}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}2\cos ^{2}x+\cos x-1&=0\\ \left (2\cos x-1 \right )\left (\cos x+1 \right )&=0\\ \cos x=\displaystyle \frac{1}{2}\: \: \color{black}\textrm{atau}\: \: &\cos x=-1\\ \cos x=\color{red}\cos 60^{\circ}=\cos \frac{1}{3}\pi \: \: &\\ \color{black}\textrm{atau}\: \: \cos x=\color{red}\cos 180^{\circ}&=\cos \pi \\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan ^{2}x-\tan x-6=0\: \: \textrm{pada}\: \: 0\leq x\leq \pi ,\\ &\textrm{maka himpunan nilai}\: \: \sin x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5} \right \} \\ \textrm{c}.&\left \{ -\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \\ \textrm{d}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5} \right \} \\ \textrm{e}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\tan ^{2}x-\tan x-6&=0\\ \left (\tan x-3 \right )\left (\tan x+2 \right )&=0\\ \tan x=3\: \: \textrm{atau}\: \: &\tan x=-2\\ \tan x=\displaystyle \frac{3}{1}\: \: \textrm{atau}\: \: &\tan x=\frac{-2}{1}\\ \sin x=\displaystyle \frac{3}{\sqrt{1^{2}+3^{2}}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{1^{2}+2^{2}}}\\ \sin x=\displaystyle \frac{3}{\sqrt{10}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{5}}\\ \sin x=\color{red}\displaystyle \frac{3}{10}\sqrt{10}\: \: \color{black}\textrm{atau}\: \: &\sin x=\color{red}\frac{2}{5}\sqrt{5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\quad 2\sin ^{2}x+3\cos x=0\\ &\textrm{dan}\quad 0^{0}\leq x\leq 180^{0},\\ &\textrm{maka nilai \textit{x} adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&30^{0}\\ \textrm{b}.&60^{0}\\ \textrm{c}.&\color{red}120^{0}\\ \textrm{d}.&150^{0}\\ \textrm{e}.&170^{0}\\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}2&\sin ^{2}x+3\cos x=0,\\ & \textrm{ingat identitas}\quad \sin ^{2}x+\cos ^{2}x=1\\ 2&\left ( 1-\cos ^{2}x \right )+3\cos x=0\\ 2&-2\cos ^{2}x+3\cos x=0,\\ & \textrm{(dikali dengan -1)} \\ 2&\cos ^{2}x-\cos x-2=0,\\ & \textrm{menjadi persamaan kuadrat dalam}\: \: \cos x\\ &\left ( \cos x-2 \right )\left ( 2\cos x+1 \right )=0,\qquad (\textrm{difaktorkan})\\ &\cos x-2=0\quad \textrm{atau}\quad 2\cos x+1=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\cos x=2}}\quad \textrm{atau}\quad \underset{\textrm{memenuhi}}{\underbrace{\cos x=-\displaystyle \frac{1}{2}}},\\ & \textrm{ingat rentang nilai cosinus adalah}:\left | \cos x \right |\leq 1\\ &\textrm{Selanjutnya pilih yang memenuhi, yaitu}\\ &\cos x=-\displaystyle \frac{1}{2}\\ &\cos x=\cos \left ( 180^{0}-60^{0} \right )\\ &\cos x=\cos 120^{0}\\ &\therefore \quad x=\color{red}120^{0} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Akar-akar dari persamaan}\\ & 4\sin ^{2}x+4\cos x=1 \quad \textrm{pada selang}\\ & -\pi \leq x\leq \pi \quad \textrm{adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{3}{2}\: \: \textrm{atau}\: \: -\frac{1}{2}\\ \textrm{b}.&-\displaystyle \frac{3}{2}\: \: \textrm{atau}\: \: \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{3}{2}\pi \: \: \textrm{atau}\: \: -\frac{1}{2}\pi \\ \textrm{d}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{atau}\: \: -\frac{1}{3}\pi \\ \textrm{e}.&\color{red}\displaystyle -\frac{2}{3}\pi \: \: \textrm{atau}\: \: \frac{2}{3}\pi \\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{e}\\ &\begin{aligned}4&\sin ^{2}x+4\cos x=1\\ 4&\left ( 1-\cos ^{2}x \right )+4\cos x-1=0\\ 4&-4\cos ^{2}x+4\cos x-1=0\\ 4&\cos ^{2}x-4\cos x-3=0\\ &\left ( 2\cos x-3 \right )\left ( 2\cos x+1 \right )=0\\ &\left ( 2\cos x-3 \right )=0\quad \textrm{atau}\quad \left ( 2\cos x+1 \right )=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\cos x=\displaystyle \frac{3}{2}}}\quad \textrm{atau} \quad\underset{\textrm{memenuhi}}{\underbrace{\cos x=-\displaystyle \frac{1}{2}}}\\ &\textrm{pilih yang memenuhi persamaan, yaitu}\\ &\cos x=-\displaystyle \frac{1}{2}\\ &\cos x=\cos \left ( 180^{0}-60^{0} \right )\\ &\cos x=\cos 120^{0}\\ &\therefore \quad x=120^{0}=\color{red}\displaystyle \frac{2}{3}\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Himpunan penyelesaian persamaan}\\ &\sin ^{2}2x+2\sin x\cos x -2=0\\ & \textrm{pada selang}\quad 0 \leq x\leq 360^{0} \quad \textrm{adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 45^{0},135^{0} \right \}\\ \textrm{b}.&\color{red}\left \{ 45^{0},225^{0} \right \}\\ \textrm{c}.&\left \{ 135^{0},180^{0} \right \}\\ \textrm{d}.&\left \{ 135^{0},225^{0} \right \} \\ \textrm{e}.&\left \{ 135^{0},315^{0} \right \} \\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{b}\\ &\begin{aligned}&\sin ^{2}2x+2\sin x\cos x -2=0\\ &\sin ^{2}2x+\sin 2x-2=0,\\ & \textrm{(ingat bahwa)}:\: \sin 2x=2\sin x\cos x\\ &\left ( \sin 2x+2 \right )\left ( \sin 2x-1 \right )=0\\ &\sin 2x+2=0\quad \textrm{atau}\quad \sin 2x-1=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\sin 2x=-2}}\quad \textrm{atau}\quad \underset{\textrm{memenuhi}}{\underbrace{\sin 2x=1}}\\ &\textrm{pilih yang memenuhi persamaan, yaitu}\\ &\sin 2x=1\\ &\sin 2x=\sin 90^{0}\\ &\quad 2x=90^{0}+k.360^{0}\quad \textrm{atau}\\ &\quad 2x=\left (180^{0}-90^{0} \right )+k.360^{0}\\ & \textrm{(cukup ambil 1 persamaan saja, karena sama)}\\ &\qquad x=45^{0}+k.180^{0}\\ &k=0\Rightarrow x=45^{0}+0.180^{0}=45^{0}\\ &k=1\Rightarrow x=45^{0}+1.180^{0}=45^{0}+180^{0}=225^{0}\\ &k=2\Rightarrow x=45^{0}+2.180^{0}=45^{0}+360^{0}=405^{0}\\ & \textrm{(tidak memenuhi rentang)}\\ &\therefore \quad \textbf{HP}=\color{red}\left \{ 45^{0},225^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 36.&\textrm{Bentuk}\: \: \sqrt{3}\cos x-\sin x\: \: \textrm{untuk}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{dapat dinyatakan sebagai}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\cos \left ( x+\displaystyle \frac{\pi }{6} \right )\\ \textrm{b}.&2\cos \left ( x+\displaystyle \frac{7\pi }{6} \right ) \\ \color{red}\textrm{c}.&2\cos \left ( x-\displaystyle \frac{11\pi }{6} \right )\\ \textrm{d}.&2\cos \left ( x-\displaystyle \frac{7\pi }{6} \right )\\ \textrm{e}.&2\cos \left ( x-\displaystyle \frac{\pi }{6} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{c}\\ &\begin{aligned}\sqrt{3}\cos x-\sin x&=k\cos (x-\alpha )\\ (1)\qquad & (a,b)=\begin{cases} a &=\sqrt{3} \\ b &=-1 \end{cases}\\ \textrm{maka}\: &\textrm{titik ada dikadran IV}\\ (2)\: \quad k&=\sqrt{a^{2}+b^{2}}\\ &=\sqrt{\sqrt{3}^{2}+(-1)^{2}}\\ &=\sqrt{4}=2\\ (3)\quad \alpha &=\arctan \frac{b}{a}=\arctan \left (\frac{-1}{\sqrt{3}} \right )=-30^{\circ}\\ &=\left ( 360^{\circ}-30^{\circ} \right )=330^{\circ}=\displaystyle \frac{11}{6}\pi \\ \textrm{sehingga}&\\ \sqrt{3}\cos x-\sin x&=\color{red}2\cos \left ( x-\displaystyle \frac{11}{6}\pi \right )\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Nilai-nilai}\: \: x\: \: \textrm{yang terletak pada}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{yang memenuhi persamaan}\: \: \sqrt{3}\cos x+\sin x=\sqrt{2}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \color{red}\textrm{b}.&75^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{c}.&15^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \textrm{d}.&15^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{e}.&15^{\circ}\: \: \textrm{atau}\: \: 75^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}\sqrt{3}\cos x+\sin x&=\sqrt{2}\\ \sqrt{\sqrt{3}^{2}+1^{2}}&\left ( \cos \left ( \alpha -\arctan \displaystyle \frac{1}{\sqrt{3}} \right ) \right )=\sqrt{2}\\ 2\cos \left ( x-30^{\circ} \right )&=\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\cos 45^{\circ}\\ \left ( x-30^{\circ} \right )&=\pm 45^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=75^{\circ}\\ x_{2}&=\color{red}-15^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=75^{\circ}+360^{\circ}=\color{red}435^{\circ}\\ x_{4}&=-15^{\circ}+360^{\circ}=\color{red}345^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Diketahui fungsi trigonometri}\: \: f(x)=\displaystyle \frac{1}{2}\sin 3x\\ &\textrm{perhatikanlah pernyataan-pernyataan berikut}\\ &(1)\quad \textrm{hasil dari}\: \: f(0)+f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &(2)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )+f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(3)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )-f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(4)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{3} \right )-f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &\textrm{Pernyataan yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(1)\: \: \textrm{dan}\: \: (2)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \textrm{c}.&(1)\: \: \textrm{dan}\: \: (4)\\ \color{red}\textrm{d}.&(2)\: \: \textrm{dan}\: \: (3)\\ \textrm{e}.&(3)\: \: \textrm{dan}\: \: (4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}\textrm{Diketahui}&\\ f(x)&=\displaystyle \frac{1}{2}\sin 3x\\ \textrm{maka}\qquad&\\ f(0)&=\displaystyle \frac{1}{2}\sin 3(0^{\circ})=0\\ f\left ( \displaystyle \frac{\pi }{3} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{3} \right )=0\\ f\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{6} \right )=\color{red}\displaystyle \frac{1}{2}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Himpunan penyelesaian dari}\: \: \sin x-\sqrt{3}\cos x=-1\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0^{\circ},120^{\circ} \right \}\\ \textrm{b}.&\left \{ 90^{\circ},330^{\circ} \right \}\\ \textrm{c}.&\left \{ 60^{\circ},180^{\circ} \right \}\\ \textrm{d}.&\left \{ 90^{\circ},120^{\circ} \right \}\\ \color{red}\textrm{e}.&\left \{ 30^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\sin x-\sqrt{3}\cos x&=-1\\ \sqrt{1^{2}+\left ( -\sqrt{3} \right )^{2}}\cos \left ( x-\alpha \right )&=-1\\ a&=x=-\sqrt{3},\: \: b=y=1\\ &\textbf{kuadran II}\\ \alpha &=\arctan \left ( \displaystyle \frac{1}{-\sqrt{3}} \right )=-30^{\circ}\\ &=180^{\circ}-30^{\circ}=150^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( x-150^{\circ} \right )&=-1\\ \cos \left ( x-150^{\circ} \right )&=\displaystyle \frac{-1}{2}\\ \cos \left ( x-150^{\circ} \right )&=\cos 120^{\circ}\\ \left ( x-150^{\circ} \right )&=\pm 120^{\circ}+k.360^{\circ}\\ x&=150^{\circ}\pm 120^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: &k=0\\ x_{1}&=270^{\circ}\\ x_{2}&=30^{\circ}\\ \textrm{saat}\: \: &k=1\\ x_{3}&=270^{\circ}+360^{\circ}=....\\ x_{4}&=\color{red}30^{\circ}+360^{\circ}=.... \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: \tan x\: \: \textrm{yang memenuhi persamaan}\\ &\cos 2x+7\cos x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{5}\sqrt{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\cos 2x+7\cos x-3&=0\\ 2\cos ^{2}x-1+7\cos x-3&=0\\ 2\cos ^{2}x+7\cos x-4&=0\\ \left ( \cos x+4 \right )\left (2 \cos -1 \right )&=0\\ \color{red}\cos x=-4\: \: \color{black}\textrm{atau}\: \: \color{red}\cos x&=\displaystyle \frac{1}{2}=\cos 60^{\circ}\\ x&=60^{\circ},\\ \textrm{maka}&\\ \tan 60^{\circ}&=\sqrt{3}\\\\ \textrm{ dan ingat bahwa}&\: \: \cos x=-4\: \: \color{red}\textrm{tidak memenuhi} \end{aligned} \end{array}$