HOTS Zone : Persamaan Trigonometri

Table of Contents

Berikut ini adalah kumpulan soal mengenai Persamaan Trigonometri. Jika ingin bertanya soal, silahkan gabung ke grup Matematika Idhamdaz.

Tipe:

STANDARSNBTHOTS

No. 1

Banyak bilangan real t yang memenuhi persamaan sin⁴ t − cos⁴ t = 1 dengan −π ≤ t ≤ π adalah ....

1. 0
2. 4
3. 8

4. 2
5. 1

Penyisihan LMNAS Ke-28 SMA

ALTERNATIF PENYELESAIAN

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

• 2t = π + 2kπ
  $t=\dfrac{\pi}2+k\pi$
  
  k = −1 ⟶ $t=-\dfrac{\pi}2$ k = 0 ⟶ $t=\dfrac{\pi}2$
• 2t = −π + 2kπ
  $t=-\dfrac{\pi}2+k\pi$
  
  k = 0 ⟶ $t=-\dfrac{\pi}2$ k = 1 ⟶ $t=\dfrac{\pi}2$

Jadi,
JAWAB:

---

No. 2

$\sqrt{a}\cos x-\sqrt{a}\sin x=\dfrac{m^2\cos2x}{\cos x+\sin x}$

Related: loading

maka $\dfrac{a}{m^4}=$ ....

Grup Whatsapp

ALTERNATIF PENYELESAIAN

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

Jadi, $\dfrac{a}{m^4}=1$.

---

No. 3

Diberikan (24 cos x)² = (24 sin x)³, dengan ${0 \lt x \lt\dfrac{\pi}2}$. Nilai dari cot² x = ....

Grup Whatsapp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}(24\cos x{)}^2& =(24\sin x{)}^3\\ {24}^2{\cos }^{2}x& ={24}^3{\sin }^{3}x\\ {\cos }^{2}x& =24{\sin }^{3}x\\ 1-{\sin }^{2}x& =24{\sin }^{3}x\\ 24{\sin }^{3}x+{\sin }^{2}x-1& =0\\ (3\sin x-1)(8{\sin }^{2}x+3\sin x+1)& =0\\ \sin x& ={\displaystyle \frac{1}{3}}\end{array}$

$\begin{array}{rl}{\cot }^{2}x& ={\displaystyle \frac{{\cos }^{2}x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{1-{\sin }^{2}x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{1}{{\sin }^{2}x}}-1\\ & ={\left({\displaystyle \frac{1}{\sin x}}\right)}^2-1\\ & ={\left(3\right)}^2-1\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black8}}}}\end{array}$

Jadi, cot² x = 8.

---