HOTS Zone : Persamaan Trigonometri

Berikut ini adalah kumpulan soal Persamaan Trigonometri tipe HOTS. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.
Tipe :

• Standar
• SBMPTN
• HOTS

No. 1

\sqrt{a}\cos x-\sqrt{a}\sin x=\dfrac{m^2\cos2x}{\cos x+\sin x}
maka \dfrac{a}{m^4}=

SUMBER

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

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

No. 2

Diberikan {(24\cos x)^2= (24 \sin x)^3}, dengan {0 \lt x \lt\dfrac12}. Nilai dari \cot^2 x =

Grup Telegram

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

\sqrt{3^2-1^2}=\sqrt8

$\begin{array}{rl}\cot x& ={\displaystyle \frac{\sqrt{8}}{1}}\\ & =\sqrt{8}\\ {\cot }^{2}x& =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, \cot^2 x =8.