4.2 Determine the general solution if $$ 2 \sin x \cos x-\cos ^{2} x=0 $$

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UpStudy AI · Accepted Answer

1. Start with the equation:
   $$
   2 \sin x \cos x - \cos^2 x = 0
   $$
   
2. Factor out the common factor $$\cos x$$:
   $$
   \cos x (2 \sin x - \cos x) = 0
   $$

3. Set each factor equal to zero:

**Case 1:** 
   $$
   \cos x = 0
   $$
   The solutions to $$\cos x = 0$$ are:
   $$
   x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}
   $$

**Case 2:**
   $$
   2 \sin x - \cos x = 0
   $$
   Solve for $$\tan x$$ (assuming $$\cos x \neq 0$$):
   $$
   2 \sin x = \cos x \quad \Rightarrow \quad \frac{2 \sin x}{\cos x} = 1 \quad \Rightarrow \quad 2 \tan x = 1
   $$
   Therefore:
   $$
   \tan x = \frac{1}{2}
   $$
   The general solution for $$\tan x = \frac{1}{2}$$ is:
   $$
   x = \arctan\left(\frac{1}{2}\right) + k\pi, \quad k \in \mathbb{Z}
   $$

4. Combine the solutions:
   $$
   \boxed{x=\frac{\pi}{2}+k\pi \quad \text{or} \quad x=\arctan\left(\frac{1}{2}\right)+k\pi,\quad k \in \mathbb{Z}}
   $$
The general solution is: $$ x = \frac{\pi}{2} + k\pi \quad \text{or} \quad x = \arctan\left(\frac{1}{2}\right) + k\pi, \quad \text{where } k \text{ is any integer.} $$

4.2 Determine the general solution if

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