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

$\begin{array}{l}\\ 21.& \text{Jika diketahui}\tan (3x+{60}^0)=\sqrt{3},\\ & \text{maka nilai}x\text{yang memenuhi persamaan}\\ & \text{tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^0\\ \text{b}.& {110}^0\\ \text{c}.& {120}^0\\ \text{d}.& {130}^0\\ \text{e}.& {230}^0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\tan (3x+{60}^0)& =\sqrt{3}\\ \tan (3x+{60}^0)& =\tan {60}^0\\ (3x+{60}^0)& ={60}^0+k{.180}^0\\ 3x& =k{.180}^0\\ x& =k{.60}^0\\ k=0\to x& =0^0\\ k=1\to x& ={60}^0\\ k=2\to x& ={120}^0\\ k=3\to x& ={180}^0\\ k=4\to x& ={240}^0\\ k=5\to x& ={300}^0\\ k=\cdots \to x& =\cdots \\ \cdots \cdots \to & =....\mathbf{\text{dst}}\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Himpunan penyelesaian dari }\\ & \sqrt{3}\tan x=-1,0^0\le x\le {360}^0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^0,{150}^0\}\\ \text{b}.& \{{150}^0,{210}^0\}\\ \text{c}.& \{{150}^0,{330}^0\}\\ \text{d}.& \{{120}^0,{300}^0\}\\ \text{e}.& \{{60}^0,{240}^0\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{3}\tan x& =-1\\ \tan x& =-{\displaystyle \frac{1}{\sqrt{3}}=-\frac{1}{3}\sqrt{3}}\\ & \text{nilai di kuadran II}\\ \tan x& =\tan {150}^0\\ x& ={150}^0+k{.180}^0\\ k=0,\Rightarrow x& ={150}^0\\ k=1,\Rightarrow x& ={330}^0\\ k=2,\Rightarrow x& ={510}^0,\text{tidak}\text{memenuhi}\text{(tm)}\\ k=3,\mathbf{\text{dst}}& =\text{(tm)}\\ \therefore \mathbf{\text{HP}}& =\{{150}^0,{330}^0\}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 24.& \text{Himpunan penyelesaian dari }\\ & \cos (x-{30}^0)=-\sin {50}^0,0^0\le x\le {360}^0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{90}^0\right\}\\ \text{b}.& \{{140}^0,{150}^0\}\\ \text{c}.& \{{170}^0,{250}^0\}\\ \text{d}.& \{{80}^0,{280}^0\}\\ \text{e}.& \{{20}^0,{340}^0\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\cos (x-{30}^0)& =-\sin {50}^0\\ \cos (x-{30}^0)& =\cos ({90}^0+{50}^0)\\ \cos (x-{30}^0)& =\cos {140}^0\\ x-{30}^0& =\pm {140}^0+k{.360}^0\\ x& ={30}^0\pm {140}^0+k{.360}^0\\ k=0,\Rightarrow x_1& ={30}^0+{140}^0={170}^0\\ \Rightarrow x_2& ={30}^0-{140}^0=-{110}^0(\text{tm})\\ k=1,\Rightarrow x_1& ={30}^0+{140}^0+{360}^0={530}^0(\text{tm})\\ \Rightarrow x_2& ={30}^0-{140}^0+{360}^0={250}^0\\ \therefore \mathbf{\text{HP}}& =\{{170}^0,{250}^0\}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Himpunan penyelesaian dari}\\ & \sin 2x={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^{\circ },{210}^{\circ }\}\\ \text{b}.& \{{60}^{\circ },{240}^{\circ }\}\\ \text{c}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ }\}\\ \text{d}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sin 2x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \sin 2x& =\sin {60}^{\circ }\\ 2x& =\{\begin{array}{ll}{60}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{60}^{\circ })& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.180}^{\circ }\\ {60}^{\circ }& +k{.180}^{\circ }\end{array}\\ \text{saat}& k=0\\ x& =\{\begin{array}{ll}{30}^{\circ }& \\ {60}^{\circ }& \end{array}\\ \text{saat}& k=1\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{1.180}^{\circ }={210}^{\circ }\\ {60}^{\circ }& +{1.180}^{\circ }={240}^{\circ }\end{array}\\ \text{saat}& k=2\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{2.180}^{\circ }={390}^{\circ }\\ {60}^{\circ }& +{2.180}^{\circ }={420}^{\circ }\end{array}\\ \text{kedua}& \text{nnya tidak memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Himpunan penyelesaian dari}\\ & \tan 2x-\sqrt{3}=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\\ \text{c}.& \{{45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{e}.& \{{15}^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ },{75}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\tan 2x& -\sqrt{3}=0\\ \tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& =60+k{.180}^{\circ }\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\\ \text{saat}& k=1\\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\\ \text{saat}& k=2\\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\\ \text{saat}& k=3\\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\\ \text{saat}& k=4\\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\\ \text{tidak}& \text{memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Himpunan penyelesaian dari}\\ & \cos 3x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {180}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{40}^{\circ },{80}^{\circ }\}\\ \text{b}.& \{{50}^{\circ },{70}^{\circ }\}\\ \text{c}.& \{{40}^{\circ },{70}^{\circ },{80}^{\circ }\}\\ \text{d}.& \{{50}^{\circ },{70}^{\circ },{170}^{\circ }\}\\ \text{e}.& \{{50}^{\circ },{80}^{\circ },{170}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos 3x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 3x& =-\cos {30}^{\circ }\\ \cos 3x& =\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 3x& =\pm {150}^{\circ }+k{.360}^{\circ }\\ x& =\pm {50}^{\circ }+k{.120}^{\circ }\\ \text{saat}& k=0\\ x& =\pm {50}^{\circ }\to x={50}^{\circ }(\text{mm})\\ \text{saat}& k=1\\ x& =\pm {50}^{\circ }+{120}^{\circ }=\{\begin{array}{ll}{170}^{\circ }& (\text{mm})\\ {70}^{\circ }& (\text{mm})\end{array}\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 30.& \text{Jika diketahui}\\ & f\left({\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}\right)=\tan x,\pi \le x\le 2\pi ,\\ & \text{maka nilai}f(3)=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& {\displaystyle \frac{\pi }{2}}\\ \text{d}.& \pi \\ \text{e}.& 2\pi \end{array}\\ & \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui bahwa}& \\ f\left({\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}\right)& =\tan x,\pi \le x\le 2\pi \\ f(3)& =....\\ \text{maka selanjutnya}& \\ {\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}& =3\\ \sqrt{{\sin }^{2}x+4}& =2\\ {\sin }^{2}x+4& =4\\ {\sin }^{2}x& =0\\ \sin x& =\{\begin{array}{ll}{x}_1=0^{\circ }& +k.2\pi \\ x_2={180}^{\circ }& +k.2\pi \end{array}\\ \tan \pi =\tan 2\pi & =0\\ \therefore f(3)& =0\end{array}\end{array}$