Exercise Zone : Persamaan Trigonometri [2]

Berikut ini adalah kumpulan soal Persamaan Trigonometri tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
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No. 11

Tentukan himpunan penyelesaian persamaan berikut, untuk 0\degree\leq x\leq360\degree
\cos2x+\sin x=1

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$\begin{array}{rl}\cos 2x+\sin x& =1\\ 1-2{\sin }^{2}x+\sin x& =1\\ -2{\sin }^{2}x+\sin x& =0\\ 2{\sin }^{2}x-\sin x& =0\\ \sin x(2\sin x-1)& =0\end{array}$

• \sin x=0
  \sin x=\sin0\degree

  $\begin{array}{rl}x& =0^{\circ }+k\cdot {360}^{\circ }\\ x& =k\cdot {360}^{\circ }\end{array}$	$\begin{array}{rl}x& ={180}^{\circ }-0^{\circ }+k\cdot {360}^{\circ }\\ x& ={180}^{\circ }+k\cdot {360}^{\circ }\end{array}$
  k=0
  x=0\degree	x=180\degree
  k=1
  x=360\degree	x=540\degree (TM)

• 2\sin x-1=0
  $\begin{array}{rl}2\sin x& =1\\ \sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\end{array}$

  x=30^\circ+k\cdot360^\circ	$\begin{array}{rl}x& ={180}^{\circ }-{30}^{\circ }+k\cdot {360}^{\circ }\\ x& ={150}^{\circ }+k\cdot {360}^{\circ }\end{array}$
  k=0
  x=30\degree	x=150\degree

Jadi, Hp = \{0\degree,30\degree,150\degree,180\degree,360\degree\}.

No. 12

Nilai \tan x yang memenuhi persamaan {\cos2x+7\cos x-3=0} adalah ....

1. \sqrt3
2. \dfrac12\sqrt3
3. \dfrac13\sqrt3

4. \dfrac12
5. \dfrac15\sqrt5

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$\begin{array}{rl}\cos 2x+7\cos x-3& =0\\ 2{\cos }^{2}x-1+7\cos x-3& =0\\ 2{\cos }^{2}x+7\cos x-4& =0\\ (2\cos x-1)(\cos x+4)& =0\end{array}$
\cos x=\dfrac12 atau \cos x=-4 (TM)
$\begin{array}{rl}x& ={60}^{\circ }\\ \tan x& =\tan {60}^{\circ }\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{3}}}}}\end{array}$

Jadi, \tan x=\sqrt3.
JAWAB: A

No. 13

\sin5x+\sin3x=\sqrt3\cos x, 0\leq x\leq360\degree

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$\begin{array}{rl}\sin 5x+\sin 3x& =\sqrt{3}\cos x\\ 2\sin \left({\displaystyle \frac{5x+3x}{2}}\right)\cos \left({\displaystyle \frac{5x-3x}{2}}\right)& =\sqrt{3}\cos x\\ 2\sin 4x\cos x& =\sqrt{3}\cos x\\ 2\sin 4x\cos x-\sqrt{3}\cos x& =0\\ \cos x(2\sin 4x-\sqrt{3})& =0\end{array}$

$\begin{array}{rl}\cos x& =0\\ \cos x& =\cos {90}^{\circ }\\ x& =\{{90}^{\circ },{270}^{\circ }\}\end{array}$

$\begin{array}{rl}2\sin 4x-\sqrt{3}& =0\\ 2\sin 4x& =\sqrt{3}\\ \sin 4x& ={\displaystyle \frac{1}{2}}\sqrt{3}\\ \sin 4x& =\sin {60}^{\circ }\\ 4x& =\{{60}^{\circ },{120}^{\circ },{420}^{\circ },{480}^{\circ },{780}^{\circ },{840}^{\circ },{1140}^{\circ },{1200}^{\circ }\}\\ x& =\{{15}^{\circ },{30}^{\circ },{105}^{\circ },{120}^{\circ },{195}^{\circ },{210}^{\circ },{285}^{\circ },{300}^{\circ }\}\end{array}$

Jadi, Hp = \left\{15^\circ,30^\circ,90^\circ,105^\circ,120^\circ,195^\circ,210^\circ,270^\circ,285^\circ,300^\circ\right\}.

No. 14

2\sqrt3\sin5x-2\cos5x=2, untuk 0\degree\leq x\leq360\degree

Grup Whatsapp

$\begin{array}{rll}2\sqrt{3}\sin 5x-2\cos 5x& =2& :2\\ \sqrt{3}\sin 5x-\cos 5x& =1\end{array}$
a=-1, b=\sqrt3

$\begin{array}{rl}k& =\sqrt{a^2+b^2}\\ & =\sqrt{(-1{)}^2+{\left(\sqrt{3}\right)}^2}\\ & =\sqrt{1+3}\\ & =\sqrt{4}\\ & =2\end{array}$

$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{b}{a}}\\ & ={\displaystyle \frac{\sqrt{3}}{-1}}\\ & =-\sqrt{3}\\ \alpha & ={120}^{\circ }\end{array}$

$\begin{array}{rl}2\cos (x-{120}^{\circ })& =1\\ \cos (x-{120}^{\circ })& ={\displaystyle \frac{1}{2}}\\ \cos (x-{120}^{\circ })& =\cos {60}^{\circ }\\ x-{120}^{\circ }& =\{-{60}^{\circ },{60}^{\circ }\}\\ x& =\{-{60}^{\circ }+{120}^{\circ },{60}^{\circ }+{120}^{\circ }\}\\ x& =\{{60}^{\circ },{180}^{\circ }\}\end{array}$

Jadi, x=\{60^\circ,180^\circ\}

No. 15

Tentukan himpunan penyelesaian dari persamaan {\sin^2x + \cos^2x=\sin \left( x +\dfrac{\pi}4\right)}, untuk 0\leq x\leq\pi.

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$\begin{array}{rl}{\sin }^{2}x+{\cos }^{2}x& =\sin (x+{\displaystyle \frac{\pi }{4}})\\ 1& =\sin (x+{\displaystyle \frac{\pi }{4}})\\ \sin (x+{\displaystyle \frac{\pi }{4}})& =1\\ x+{\displaystyle \frac{\pi }{4}}& ={\displaystyle \frac{\pi }{2}}\\ x& ={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{\pi }{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\pi }{4}}}}}}\end{array}$

Jadi, Hp = \left\{\dfrac{\pi}4\right\}.

No. 16

Tentukan himpunan penyelesaian dari persamaan {-\sin\theta+\sqrt3\cos\theta=2}, untuk {0^\circ\leq x\leq360^\circ}

Tanya MTK Bot

$\begin{array}{rl}-\sin \theta +\sqrt{3}\cos \theta & =2\\ \sqrt{3}\cos \theta -\sin \theta & =2\end{array}$
a=\sqrt3, b=-1
\left(\sqrt3,-1\right) berada di kuadran IV

k=\sqrt{a^2+b^2}=\sqrt{\left(\sqrt3\right)^3+(-1)^2}=\sqrt{3+1}=\sqrt4=2

$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{b}{a}}\\ & ={\displaystyle \frac{-1}{\sqrt{3}}}\\ & =-{\displaystyle \frac{1}{3}}\sqrt{3}\\ \alpha & ={360}^{\circ }-{30}^{\circ }\\ & ={330}^{\circ }\end{array}$

$\begin{array}{rl}\sqrt{3}\cos \theta -\sin \theta & =2\\ k\cos (\theta -\alpha )& =2\\ 2\cos (\theta -{30}^{\circ })& =2\\ \cos (\theta -{30}^{\circ })& =1\\ \cos (\theta -{30}^{\circ })& =\cos 0^{\circ }\end{array}$

$\begin{array}{rl}\theta -{30}^{\circ }& =0^{\circ }+k\cdot {360}^{\circ }\\ \theta & ={30}^{\circ }+k\cdot {360}^{\circ }\end{array}$	$\begin{array}{rl}\theta -{30}^{\circ }& =-0^{\circ }+k\cdot {360}^{\circ }\\ \theta & ={30}^{\circ }+k\cdot {360}^{\circ }\end{array}$
k=0
\theta=30^\circ
k=1
\theta=390^\circ (TM)

Jadi, Hp = \{30^\circ\}.

No. 17

Tentukan himpunan penyelesaian {\sqrt6\sin x+\sqrt3=0}, untuk 0\degree\leq x\leq 360\degree

Grup Telegram

$\begin{array}{rl}\sqrt{6}\sin x+\sqrt{3}& =0\\ \sqrt{6}\sin x& =-\sqrt{3}\\ \sin x& ={\displaystyle \frac{-\sqrt{3}}{\sqrt{6}}}\cdot {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}\\ & ={\displaystyle \frac{-\sqrt{18}}{6}}\\ & ={\displaystyle \frac{-3\sqrt{2}}{6}}\\ & =-{\displaystyle \frac{1}{2}}\sqrt{2}\\ & =\sin ({180}^{\circ }+45\circ )\\ & =\sin {225}^{\circ }\end{array}$

x=225^\circ+k\cdot360^\circ	$\begin{array}{rl}x& ={180}^{\circ }-{225}^{\circ }+k\cdot {360}^{\circ }\\ x& =-{45}^{\circ }+k\cdot {360}^{\circ }\end{array}$
k=0
x=225\degree	x=-45\degree (TM)
k=1
x=585\degree (TM)	x=315\degree

Jadi, Hp = \{225\degree,315\degree\}.
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