Soal yang Akan Dibahas
Jika $ x_1 $ dan $ x_2 $ adalah solusi dari
$ \frac{2\sin x . \cos 2x}{\cos x . \sin 2x} - 5\tan x + 5 = 0 $ ,
maka $ \tan (x_1 + x_2) = .... $
A). $ -\frac{5}{7} \, $ B). $ -\frac{5}{3} \, $ C). $ \frac{\sqrt{5}}{7} \, $ D). $ \frac{\sqrt{5}}{3} \, $ E). $ \frac{5}{3} \, $
A). $ -\frac{5}{7} \, $ B). $ -\frac{5}{3} \, $ C). $ \frac{\sqrt{5}}{7} \, $ D). $ \frac{\sqrt{5}}{3} \, $ E). $ \frac{5}{3} \, $
$\spadesuit $ Konsep Dasar
*). Rumus-rumus dasar trigonometri :
$ \sin 2x = 2\sin x . \cos x $
$ \cos 2x = \cos ^2 x - \sin ^2 x = (\cos x - \sin x)(\cos x + \sin x) $
$ \tan x = \frac{\sin x}{\cos x } $
$ \tan (x+y) = \frac{\tan x + \tan y}{1 - \tan x . \tan y } $
*). Bentuk pecahan : $ a = nb \rightarrow \frac{a}{b} = n $
*). Rumus-rumus dasar trigonometri :
$ \sin 2x = 2\sin x . \cos x $
$ \cos 2x = \cos ^2 x - \sin ^2 x = (\cos x - \sin x)(\cos x + \sin x) $
$ \tan x = \frac{\sin x}{\cos x } $
$ \tan (x+y) = \frac{\tan x + \tan y}{1 - \tan x . \tan y } $
*). Bentuk pecahan : $ a = nb \rightarrow \frac{a}{b} = n $
$\clubsuit $ Pembahasan
*). Menentukan nilai $ \tan x_1 $ dan $ \tan x_2 $ :
$\begin{align} \frac{2\sin x . \cos 2x}{\cos x . \sin 2x} - 5\tan x + 5 & = 0 \\ \frac{2\sin x . \cos 2x}{\cos x . 2\sin x . \cos x} - 5\frac{\sin x}{\cos x } + 5 & = 0 \\ \frac{\cos 2x}{\cos ^2 x } - 5\frac{\sin x}{\cos x } + 5 & = 0 \, \, \, \, \, \text{(kalikan } \cos ^2 x) \\ \cos 2x - 5\sin x \cos x + 5\cos ^2 x & = 0 \\ \cos 2x + 5 \cos x ( \cos x - \sin x ) & = 0 \\ (\cos x - \sin x)(\cos x + \sin x) + 5 \cos x & ( \cos x - \sin x ) = 0 \\ (\cos x - \sin x)(\cos x + \sin x + 5 \cos x ) & = 0 \\ (\cos x - \sin x)(6\cos x + \sin x ) & = 0 \\ \sin x = \cos x \vee \sin x & = - 6\cos x \\ \frac{\sin x}{\cos x} = 1 \vee \frac{\sin x}{\cos x} & = - 6 \\ \tan x = 1 \vee \tan x & = - 6 \\ \tan x_1 = 1 \vee \tan x_2 & = - 6 \end{align} $
*). Menentukan nilai $ \tan ( x_1 + x_2) $ :
$\begin{align} \tan (x_1 + x_2) & = \frac{\tan x_1 + \tan x_2}{1 - \tan x_1 . \tan x_2 } \\ & = \frac{1 + (-6)}{1 - 1 . (-6) } \\ & = \frac{-5}{1 + 6 } = -\frac{5}{7} \end{align} $
Jadi, nilai $ \tan (x_1 + x_2) = -\frac{5}{7} . \, \heartsuit $
*). Menentukan nilai $ \tan x_1 $ dan $ \tan x_2 $ :
$\begin{align} \frac{2\sin x . \cos 2x}{\cos x . \sin 2x} - 5\tan x + 5 & = 0 \\ \frac{2\sin x . \cos 2x}{\cos x . 2\sin x . \cos x} - 5\frac{\sin x}{\cos x } + 5 & = 0 \\ \frac{\cos 2x}{\cos ^2 x } - 5\frac{\sin x}{\cos x } + 5 & = 0 \, \, \, \, \, \text{(kalikan } \cos ^2 x) \\ \cos 2x - 5\sin x \cos x + 5\cos ^2 x & = 0 \\ \cos 2x + 5 \cos x ( \cos x - \sin x ) & = 0 \\ (\cos x - \sin x)(\cos x + \sin x) + 5 \cos x & ( \cos x - \sin x ) = 0 \\ (\cos x - \sin x)(\cos x + \sin x + 5 \cos x ) & = 0 \\ (\cos x - \sin x)(6\cos x + \sin x ) & = 0 \\ \sin x = \cos x \vee \sin x & = - 6\cos x \\ \frac{\sin x}{\cos x} = 1 \vee \frac{\sin x}{\cos x} & = - 6 \\ \tan x = 1 \vee \tan x & = - 6 \\ \tan x_1 = 1 \vee \tan x_2 & = - 6 \end{align} $
*). Menentukan nilai $ \tan ( x_1 + x_2) $ :
$\begin{align} \tan (x_1 + x_2) & = \frac{\tan x_1 + \tan x_2}{1 - \tan x_1 . \tan x_2 } \\ & = \frac{1 + (-6)}{1 - 1 . (-6) } \\ & = \frac{-5}{1 + 6 } = -\frac{5}{7} \end{align} $
Jadi, nilai $ \tan (x_1 + x_2) = -\frac{5}{7} . \, \heartsuit $
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