Contoh Soal 3 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 11.& \text{Himpunan penyelesaian persamaan}\\ & 3\cos 2x+5\sin x+1=0\text{untuk}{0}^{\circ }\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{b}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{c}.& \left\{{\displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi }\right\}\\ \text{d}.& \left\{{\displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi }\right\}\\ \text{e}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi }\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}3\cos 2x+5\sin x+1& =0\\ 3(1-2{\sin }^{2}x)+5\sin x+1& =0\\ -6{\sin }^2+5\sin x+4& =0\\ 6{\sin }^{2}x-5\sin x-4& =0\\ (3\sin x-4)(2\sin x+1)& =0\\ \sin x={\displaystyle \frac{4}{3}\text{atau}\sin x}& =-\frac{1}{2}\\ \sin x& =\sin {150}^{\circ }=\frac{5}{6}\pi \\ x& =\{\begin{array}{ll}{\displaystyle \frac{7}{6}\pi }& +k.2\pi \\ \pi -{\displaystyle \frac{7}{6}\pi }& +k.2\pi \end{array}\\ \text{saat}k& =0\\ x_1& ={\displaystyle \frac{7}{6}\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi }\\ \text{saat}k& =1\\ x_1& ={\displaystyle \frac{7}{6}\pi +2\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi +2\pi =\frac{11}{6}\pi }\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Himpunan penyelesaian dari}\\ & \sqrt{3}\sin 2x+2{\cos }^{2}x=-1\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{240}^{\circ },{300}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{60}^{\circ }\}\\ \text{c}.& \{{150}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{120}^{\circ },{300}^{\circ }\}\\ \text{e}.& \{{60}^{\circ },{150}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3}\sin 2x+2{\cos }^{2}x& =-1\\ \sqrt{3}\sin 2x+1+\cos 2x& =-1\\ \sqrt{3}\sin 2x+\cos 2x& =-2\\ \sqrt{{\sqrt{3}}^2+1^2}\cos (2x-\alpha )& =-2\\ a=x=1,b& =y=\sqrt{3}\\ \alpha & =\arctan {\displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}}\\ \alpha & ={60}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (2x-{60}^{\circ })& =-2\\ \cos (2x-{60}^{\circ })& =-1\\ \cos (2x-{60}^{\circ })& =\cos {180}^{\circ }\\ (2x-{60}^{\circ })& =\pm {180}^{\circ }+k{.360}^{\circ }\\ 2x& ={60}^{\circ }\pm {180}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {90}^{\circ }+k{.180}^{\circ }\\ \text{saat}k& =0\\ x_1& ={120}^{\circ }\\ x_2& =-{60}^{\circ }\\ \text{saat}k& =1\\ x_3& ={120}^{\circ }+{360}^{\circ }=....\\ x_2& =-{60}^{\circ }+{360}^{\circ }={300}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jika}3\sin \theta +4\cos \theta =5\text{maka}\\ & \text{nilai dari}\sin \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0,3\\ \text{b}.& 0,60\\ \text{c}.& 0,75\\ \text{d}.& 0,80\\ \text{e}.& 1,20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}3\sin \theta +4\cos \theta & =5\\ k\cos (\theta -\alpha )& =5\\ a=x=4,& b=y=3\\ \theta & =\arctan {\displaystyle \frac{b}{a}}\\ \theta & =\arctan {\displaystyle \frac{3}{4}}\\ \text{atau}& \\ \tan \theta & ={\displaystyle \frac{3}{4}}\\ \text{maka}& \\ \sin \theta & ={\displaystyle \frac{3}{\sqrt{3^2+4^2}}}\\ & ={\displaystyle \frac{3}{5}}\\ & =0,6\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Jika}\tan \theta +\sec \theta =x\text{maka}\\ & \text{nilai dari}\tan \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2x}{x^2-1}}\\ \text{b}.& {\displaystyle \frac{2x}{x^2+1}}\\ \text{c}.& {\displaystyle \frac{x^2+1}{2x}}\\ \text{d}.& {\displaystyle \frac{x^2-1}{2x}}\\ \text{e}.& {\displaystyle \frac{x^2-1}{x^2+1}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Langkah}& 1\\ \tan \theta +\sec \theta & =x\\ {\displaystyle \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}& =x\\ {\displaystyle \frac{\sin \theta +1}{\cos \theta }}& =x\\ \sin \theta +1& =x\cos \theta ...........1\\ \text{Langkah}& 2\\ {(\tan \theta +\sec \theta )}^2& =x^2\\ {\tan }^2\theta +2\tan \theta \sec \theta +{\sec }^2\theta & =x^2\\ {\sec }^2\theta -1+2\tan \theta \sec \theta & +{\sec }^2\theta =x^2\\ 2{\sec }^2\theta +2\tan \theta \sec \theta & =x^2+1\\ {\displaystyle \frac{2}{{\cos }^2\theta }+\frac{2\sin \theta }{{\cos }^2\theta }}& =x^2+1\\ 1+\sin \theta & =\left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2\theta .....2\\ \text{Langkah}& 3\\ \left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2& =x\cos \theta \\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \text{maka}& (\mathbf{\text{dengan sisi segitiga}})\\ \tan \theta ={\displaystyle \frac{\sqrt{{(x^2+1)}^2-(2x{)}^2}}{2x}}& ={\displaystyle \frac{\sqrt{x^4+2x^2+1-4x^2}}{2x}}\\ \tan \theta & ={\displaystyle \frac{\sqrt{x^4-2x^2+1}}{2x}}\\ & ={\displaystyle \frac{\sqrt{{(x^2-1)}^2}}{2x}}\\ & ={\displaystyle \frac{x^2-1}{2x}}\end{array}\end{array}$

$\begin{array}{rl}.\mathbf{\text{Sehingga}}& \mathbf{\text{dari kasus di atas didapatkan}}\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1}}\\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \tan \theta & ={\displaystyle \frac{x^2-1}{2x}}\end{array}$

$\begin{array}{l}\\ 15.& \text{Jika}\sec x+\tan x={\displaystyle \frac{3}{2}\text{untuk}}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\sin x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{13}}\\ \text{b}.& {\displaystyle \frac{12}{13}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle \frac{2}{13}}\\ \text{e}.& {\displaystyle \frac{5}{12}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sec x+\tan x& ={\displaystyle \frac{3}{2}}\\ \text{ingat saat}& \text{mengerjakan soal no}.14\\ \sec \theta +\tan \theta & =x\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1},\text{maka}}\\ \sin x& ={\displaystyle \frac{{\left({\displaystyle \frac{3}{2}}\right)}^2-1}{{\left({\displaystyle \frac{3}{2}}\right)}^2+1}}\\ & ={\displaystyle \frac{{\displaystyle \frac{9}{4}-1}}{{\displaystyle \frac{9}{4}+1}}}\\ & =\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{13}{4}}}\\ & ={\displaystyle \frac{5}{13}}\end{array}\end{array}$