Lanjutan 3 Persamaan Trigonometri

 B. 1 Persamaan Trigonometri Sederhana

Dalam penyelesaian persamaan trigonometri sederhana dapat digunakan salah satu rumus berikut, yaitu:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=({180}^{\circ }-{\alpha }^{\circ })+k{.360}^{\circ }\end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=-{\alpha }^{\circ }+k{.360}^{\circ }\end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k{.180}^{\circ }\end{array}$.

Jika sudutnya dinyatakan dalam phi radian $(\pi \text{dibaca}:phi)$, maka persamaan trigonometri sederhananya adalah:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=(\pi -{\alpha }^{\circ })+k.2\pi \end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=-{\alpha }^{\circ }+k.2\pi \end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k.\pi \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah akar-akar persamaan trigonometri}\\ & \text{berikut dan tentukan pula himpunan}\\ & \text{penyelesaiannya untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{a}.\sin x=\sin {50}^{\circ }\\ & \text{b}.\cos x=\cos {50}^{\circ }\\ & \text{c}.\tan x=\tan {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}.\text{a}.\sin x& =\sin {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{50}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ {130}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{50}^{\circ },{130}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{b}.\cos x& =\cos {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ -{50}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ -{50}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{50}^{\circ }+{360}^{\circ }={410}^{\circ }(\text{tidak memenuhi})& \\ -{50}^{\circ }+{360}^{\circ }={310}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{50}^{\circ },{310}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{c}.\tan x& =\tan {50}^{\circ }\\ x& ={50}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={50}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={50}^{\circ }+{180}^{\circ }={230}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={50}^{\circ }+{360}^{\circ }={410}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{50}^{\circ },{230}^{\circ }\}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari }\\ & \text{persamaan-persamaan trigonometri berikut}\\ & \text{ini untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \begin{array}{l}\\ \text{a}.& \sin x={\displaystyle \frac{1}{2}}& \text{f}.& \tan x=-{\displaystyle \frac{1}{3}\sqrt{3}}& \text{k}.& \sin 2x={\displaystyle \frac{1}{2}}\\ \text{b}.& \cos x={\displaystyle \frac{1}{2}\sqrt{3}}& \text{g}& 2\cos x=-\sqrt{3}& \text{l}.& \cos 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& \tan x=\sqrt{3}& \text{h}& 3\tan x=\sqrt{3}& \text{m}.& \tan 2x=\sqrt{3}\\ \text{d}.& \sin x=-1& \text{i}.& \sin x=\sin {46}^{\circ }& \text{n}.& \sin (2x-{30}^{\circ })=\sin {45}^{\circ }\\ \text{e}.& \cos x=-{\displaystyle \frac{1}{2}\sqrt{2}}& \text{j}.& \cos x=\cos {93}^{\circ }& \text{o}.& \sin (2x+{60}^{\circ })=\sin {90}^{\circ }\end{array}\end{array}$

$.\text{Jawab}:$

$\begin{array}{rl}.\text{a}.\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ {150}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{30}^{\circ },{150}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{b}.\cos x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos x& =\cos {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ -{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ -{30}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tidak memenuhi})& \\ -{30}^{\circ }+{360}^{\circ }={330}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{30}^{\circ },{330}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{c}.\tan x& =\sqrt{3}\\ \tan x& =\tan {60}^{\circ }\\ x& ={60}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={60}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={60}^{\circ }+{180}^{\circ }={240}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={60}^{\circ }+{360}^{\circ }={420}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{60}^{\circ },{240}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{d}.\sin x& =-1\\ \sin x& =\sin {270}^{\circ }\\ x& =\{\begin{array}{ll}{270}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{270}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}{270}^{\circ }& \text{memenuhi}\\ -{90}^{\circ }& \text{tidak memenuhi}\end{array}\\ k=1& \text{tidak memenuhi semuanya}\\ \text{HP}& =\left\{{270}^{\circ }\right\}\end{array}$.

$\begin{array}{rl}.\text{k}.\sin 2x& ={\displaystyle \frac{1}{2}}\\ \sin 2x& =\sin {30}^{\circ }\\ 2x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{15}^{\circ }& +k{.180}^{\circ }\\ ({90}^{\circ }-{15}^{\circ })& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }+{180}^{\circ }={195}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{75}^{\circ },{195}^{\circ },{255}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{l}.\cos 2x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 2x& =-\cos {30}^{\circ }=\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 2x& =\{\begin{array}{ll}{150}^{\circ }& +k{.360}^{\circ }\\ -{150}^{\circ }& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{75}^{\circ }& +k{.180}^{\circ }\\ -{75}^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{75}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }+{180}^{\circ }={105}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{360}^{\circ }={435}^{\circ }(\text{tidak memenuhi})& \\ -{75}^{\circ }+{360}^{\circ }={285}^{\circ }(\text{memenuhi})& \end{array}\\ k=3& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{75}^{\circ },{105}^{\circ },{255}^{\circ },{285}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{m}.\tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& ={60}^{\circ }+k{.180}^{\circ }\\ \text{sehin}& \text{gga}\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={30}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\text{memenuhi}\\ k=3& \\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\text{memenuhi}\\ k=4& \\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{n}.\sin (2x-{30}^{\circ })& =\sin {45}^{\circ }\\ (2x-{30}^{\circ })& =\{\begin{array}{ll}{45}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{45}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{45}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\\ {135}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}37,5^{\circ }& +k{.180}^{\circ }\\ 82,5^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }& \\ 82,5^{\circ }& \end{array}\\ k=1& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }+{180}^{\circ }& =217,5^{\circ }\\ 82,5^{\circ }+{180}^{\circ }& =262,5^{\circ }\end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{37,5^{\circ },82,5^{\circ },217,5^{\circ },262,5^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{o}.\sin (2x+{60}^{\circ })& =\sin {90}^{\circ }\\ (2x+{60}^{\circ })& =\{\begin{array}{ll}{90}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{90}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\\ {90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& ={15}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}\\ x& ={15}^{\circ }\\ k=1& \text{diperoleh}\\ x& ={15}^{\circ }+{180}^{\circ }={195}^{\circ }\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{195}^{\circ }\}\end{array}$