Given the equation (3 \cdot \sin x - 4 \cdot \cos x = 5) and (\frac{\pi}{2} < x < \pi), find (\tan x).
To solve this problem, we can use a trigonometric identity and transformation. The equation is of the form (a \cdot \sin x + b \cdot \cos x = c).
First, find (r) where:
r = \sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
Thus, the equation can be rewritten using the amplitude (r):
r(\sin x \cdot \frac{a}{r} - \cos x \cdot \frac{b}{r}) = c
Given:
a = 3, \quad b = -4
\sin x \cdot \frac{3}{5} - \cos x \cdot \frac{-4}{5} = 1
This simplifies to:
\sin(x - \alpha) = \frac{1}{1}
To find (\tan x), use the identity:
\tan(x) = \frac{\sin(x)}{\cos(x)}
Since (\frac{\pi}{2} < x < \pi), this means (\sin x) is positive and (\cos x) is negative. Therefore, (\tan x) should be negative.
To determine (\alpha), use:
\cos \alpha = \frac{3}{5}, \quad \sin \alpha = \frac{4}{5}
Now, calculate (\tan x):
\tan(\alpha) = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{3/5} = \frac{4}{3}
So, the solution:
-\tan(x) = -\frac{4}{3}
Thus, the correct answer choice is B) -\frac{4}{3}
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