Question

The general solution satisfying both the equations \(\sin x = -\frac{3}{5}\) and \(\cos x = -\frac{4}{5}\) is:

• A. \(x = (2n+1)\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right), n \in Z \)
• B. \(x = 2n\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right), n \in Z \)
• C. \(x = n\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right), n \in Z \)
• D. \(x = n\pi \pm \operatorname{Tan}^{-1}\left(\frac{3}{4}\right), n \in Z \)

Hint:

When given both \(\sin x\) and \(\cos x\) values, first determine the correct quadrant for \(x\). Then, find \(\tan x\). While the general solution for \(\tan x = \tan \alpha\) is \(x = n\pi + \alpha\), you must choose the appropriate form of \(n\) (even or odd) to ensure \(x\) lies in the correct quadrant as determined by the signs of \(\sin x\) and \(\cos x\).

Answer 1

The Correct Option is A

Step 1: Determine the quadrant of \( x \).
We are given two conditions:
1. \( \sin x = -\frac{3}{5} \) (sine is negative)
2. \( \cos x = -\frac{4}{5} \) (cosine is negative)
For both sine and cosine to be negative, the angle \( x \) must lie in the third quadrant.
Step 2: Find the value of \( \tan x \).
We can find \( \tan x \) using the given values of \( \sin x \) and \( \cos x \):
\[ \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4} \] Step 3: Write the general solution for \( \tan x = \frac{3}{4} \).
Let \( \alpha = \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \). By definition of the inverse tangent function, \( \alpha \) is an angle in the first quadrant (\( 0 < \alpha < \frac{\pi}{2} \)) such that \( \tan \alpha = \frac{3}{4} \).
The general solution for \( \tan x = \tan \alpha \) is given by: \[ x = n\pi + \alpha, \quad \text{where } n \in \mathbb{Z} \] Substituting \( \alpha \): \[ x = n\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \] Step 4: Filter the general solution to ensure \( x \) is in the third quadrant.
We know from Step 1 that \( x \) must be in the third quadrant.
Let's test values of \( n \):
If \( n \) is an even integer (e.g., \( n = 2k \) for \( k \in \mathbb{Z} \)):
\[ x = 2k\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \] This angle lies in the first quadrant (since \( \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \) is in the first quadrant and adding \( 2k\pi \) does not change the quadrant). In the first quadrant, sine and cosine are positive, which contradicts the given conditions. If \( n \) is an odd integer (e.g., \( n = 2k+1 \) for \( k \in \mathbb{Z} \)):
\[ x = (2k+1)\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \] For \( k = 0 \), \( x = \pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \). Adding \( \pi \) to a first-quadrant angle places it in the third quadrant. In the third quadrant, sine and cosine are both negative, which is consistent with the given conditions.
This form covers all angles in the third quadrant that have a tangent of \( \frac{3}{4} \). 
Therefore, the general solution satisfying both conditions is \( x = (2n+1)\pi + \operatorname{Tan}^{-1}\left(\frac{3}{4}\right) \), where \( n \in \mathbb{Z} \).