HOTS Zone : Persamaan Trigonometri

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

Tipe:


No.

Banyak bilangan real t yang memenuhi persamaan sin4 t − cos4 t = 1 dengan −π ≤ t ≤ π adalah ....
  1. 0
  2. 4
  3. 8
  1. 2
  2. 1
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \sin^4t-\cos^4t&=1\\ \left(\sin^2t+\cos^2t\right)\left(\sin^2t-\cos^2t\right)&=1\\ \sin^2t-\cos^2t&=1\\ \cos^2t-\sin^2t&=-1\\ \cos2t&=-1\\ \cos2t&=\cos\pi \end{aligned}\)
  • 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.

$\sqrt{a}\cos x-\sqrt{a}\sin x=\dfrac{m^2\cos2x}{\cos x+\sin x}$
maka $\dfrac{a}{m^4}=$ ....
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \sqrt{a}\cos x-\sqrt{a}\sin x&=\dfrac{m^2\cos2x}{\cos x+\sin x}\\[8pt] \sqrt{a}(\cos x-\sin x)&=\dfrac{m^2\cos2x}{\cos x+\sin x}\\[8pt] \sqrt{a}(\cos x-\sin x)(\cos x+\sin x)&=m^2\cos2x\\ \sqrt{a}\left(\cos^2 x-\sin^2 x\right)&=m^2\cos2x\\ \sqrt{a}\cos2x&=m^2\cos2x\\ \sqrt{a}&=m^2\\ a&=m^4\\ \dfrac{a}{m^4}&=\boxed{\boxed{1}} \end{aligned}\)
Jadi, $\dfrac{a}{m^4}=1$.

No.

Diberikan (24 cos x)2 = (24 sin x)3, dengan ${0 \lt x \lt\dfrac{\pi}2}$. Nilai dari cot2 x = ....
ALTERNATIF PENYELESAIAN
\(\eqalign{ (24\cos x)^2&= (24 \sin x)^3\\ 24^2\cos^2x&=24^3\sin^3x\\ \cos^2x&=24\sin^3x\\ 1-\sin^2x&=24\sin^3x\\ 24\sin^3x+\sin^2x-1&=0\\ (3\sin x-1)\left(8\sin^2x+3\sin x+1\right)&=0\\ \sin x&=\dfrac13 }\)

\(\begin{aligned} \cot^2x&=\dfrac{\cos^2x}{\sin^2x}\\[4pt] &=\dfrac{1-\sin^2x}{\sin^2x}\\[4pt] &=\dfrac1{\sin^2x}-1\\[4pt] &=\left(\dfrac1{\sin x}\right)^2-1\\[4pt] &=\left(3\right)^2-1\\ &=\color{blue}\boxed{\boxed{\color{black}8}} \end{aligned}\)
Jadi, cot2 x = 8.