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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\(\eqalign{ \cos2x+\sin x&=1\\ 1-2\sin^2x+\sin x&=1\\ -2\sin^2x+\sin x&=0\\ 2\sin^2x-\sin x&=0\\ \sin x(2\sin x-1)&=0 }\)

• \sin x=0
  \sin x=\sin0\degree

  \(\eqalign{ x&=0^\circ+k\cdot360^\circ\\ x&=k\cdot360^\circ }\)	\(\eqalign{ x&=180^\circ-0^\circ+k\cdot360^\circ\\ x&=180^\circ+k\cdot360^\circ }\)
  k=0
  x=0\degree	x=180\degree
  k=1
  x=360\degree	x=540\degree (TM)

• 2\sin x-1=0
  \(\eqalign{ 2\sin x&=1\\ \sin x&=\dfrac12\\ \sin x&=\sin30^\circ }\)

  x=30^\circ+k\cdot360^\circ	\(\eqalign{ x&=180^\circ-30^\circ+k\cdot360^\circ\\ x&=150^\circ+k\cdot360^\circ }\)
  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

Grup Telegram

\(\eqalign{ \cos2x+7\cos x-3&=0\\ 2\cos^2x-1+7\cos x-3&=0\\ 2\cos^2x+7\cos x-4&=0\\ (2\cos x-1)(\cos x+4)&=0 }\)
\cos x=\dfrac12 atau \cos x=-4 (TM)
\(\eqalign{ x&=60^\circ\\ \tan x&=\tan60^\circ\\ &=\boxed{\boxed{\sqrt3}} }\)

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

No. 13

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

Grup Telegram

\(\eqalign{ \sin5x+\sin3x&=\sqrt3\cos x\\ 2\sin\left(\dfrac{5x+3x}2\right)\cos\left(\dfrac{5x-3x}2\right)&=\sqrt3\cos x\\ 2\sin4x\cos x&=\sqrt3\cos x\\ 2\sin4x\cos x-\sqrt3\cos x&=0\\ \cos x\left(2\sin4x-\sqrt3\right)&=0 }\)

\(\eqalign{ \cos x&=0\\ \cos x&=\cos90^\circ\\ x&=\left\{90^\circ,270^\circ\right\} }\)

\(\eqalign{ 2\sin4x-\sqrt3&=0\\ 2\sin4x&=\sqrt3\\ \sin4x&=\dfrac12\sqrt3\\ \sin4x&=\sin60^\circ\\ 4x&=\left\{60^\circ,120^\circ,420^\circ,480^\circ,780^\circ,840^\circ,1140^\circ,1200^\circ\right\}\\ x&=\left\{15^\circ,30^\circ,105^\circ,120^\circ,195^\circ,210^\circ,285^\circ,300^\circ\right\} }\)

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

\(\eqalign{ 2\sqrt3\sin5x-2\cos5x&=2\qquad&{\color{red}:2}\\ \sqrt3\sin5x-\cos5x&=1 }\)
a=-1, b=\sqrt3

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

\(\eqalign{ \tan\alpha&=\dfrac{b}a\\ &=\dfrac{\sqrt3}{-1}\\ &=-\sqrt3\\ \alpha&=120^\circ }\)

\(\eqalign{2\cos\left(x-120^\circ\right)&=1\\ \cos\left(x-120^\circ\right)&=\dfrac12\\ \cos\left(x-120^\circ\right)&=\cos60^\circ\\ x-120^\circ&=\{-60^\circ,60^\circ\}\\ x&=\{-60^\circ+120^\circ,60^\circ+120^\circ\}\\ x&=\{60^\circ,180^\circ\} }\)

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.

Grup Telegram

\(\eqalign{ \sin^2x + \cos^2x&=\sin \left( x +\dfrac{\pi}4\right)\\ 1&=\sin \left( x +\dfrac{\pi}4\right)\\ \sin \left( x +\dfrac{\pi}4\right)&=1\\ x +\dfrac{\pi}4&=\dfrac{\pi}2\\ x&=\dfrac{\pi}2-\dfrac{\pi}4\\ &=\boxed{\boxed{\dfrac{\pi}4}} }\)

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

\(\eqalign{ -\sin\theta+\sqrt3\cos\theta&=2\\ \sqrt3\cos\theta-\sin\theta&=2 }\)
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

\(\eqalign{ \tan\alpha&=\dfrac{b}a\\ &=\dfrac{-1}{\sqrt3}\\ &=-\dfrac13\sqrt3\\ \alpha&=360^\circ-30^\circ\\ &=330^\circ }\)

\(\eqalign{ \sqrt3\cos\theta-\sin\theta&=2\\ k\cos(\theta-\alpha)&=2\\ 2\cos(\theta-30^\circ)&=2\\ \cos(\theta-30^\circ)&=1\\ \cos(\theta-30^\circ)&=\cos0^\circ }\)

\(\eqalign{ \theta-30^\circ&=0^\circ+k\cdot360^\circ\\ \theta&=30^\circ+k\cdot360^\circ }\)	\(\eqalign{ \theta-30^\circ&=-0^\circ+k\cdot360^\circ\\ \theta&=30^\circ+k\cdot360^\circ }\)
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

\(\eqalign{ \sqrt6\sin x+\sqrt3&=0\\ \sqrt6\sin x&=-\sqrt3\\ \sin x&=\dfrac{-\sqrt3}{\sqrt6}{\color{red}\cdot\dfrac{\sqrt6}{\sqrt6}}\\ &=\dfrac{-\sqrt{18}}6\\ &=\dfrac{-3\sqrt2}6\\ &=-\dfrac12\sqrt2\\ &=\sin(180^\circ+45\circ)\\ &=\sin225^\circ }\)

x=225^\circ+k\cdot360^\circ	\(\eqalign{ x&=180^\circ-225^\circ+k\cdot360^\circ\\ x&=-45^\circ+k\cdot360^\circ\\ }\)
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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