2) $$ 5 \cos x+3=0 $$ $$ 5 \cos x=0-3 $$ $$ 5 \sin x=-3 $$ 5 $$ \cos x=\frac{-3}{5} $$ $$ x^{2}+y^{2}=r^{2} $$ $$ (-3)^{2}+y^{2}=5^{2} $$ $$ 9+y^{2}=25 $$ $$ y^{2}=25-9 $$ $$ y^{2}=16 $$ $$ y=\sqrt{16} $$ $$ y=4 $$

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UpStudy AI · Accepted Answer

1. Begin with the equation  
   $$
   5\cos x + 3 = 0.
   $$

2. Isolate the cosine term by subtracting 3 from both sides:  
   $$
   5\cos x = -3.
   $$

3. Divide both sides by 5:  
   $$
   \cos x = -\frac{3}{5}.
   $$

4. The general solutions for an equation of the form  
   $$
   \cos x = a
   $$  
   are given by  
   $$
   x = \arccos(a) + 2\pi k \quad \text{or} \quad x = -\arccos(a) + 2\pi k,
   $$  
   where $$ k $$ is any integer.

Thus, for  
   $$
   \cos x = -\frac{3}{5},
   $$  
   the solutions are  
   $$
   x = \arccos\left(-\frac{3}{5}\right) + 2\pi k \quad \text{or} \quad x = -\arccos\left(-\frac{3}{5}\right) + 2\pi k, \quad k \in \mathbb{Z}.
   $$

5. It appears that the subsequent steps involve the equation of a circle,  
   $$
   x^2 + y^2 = r^2.
   $$
   If we interpret the value $$-3$$ as the $$x$$-coordinate on a circle with radius $$ r = 5 $$, then substitute $$ x = -3 $$ and $$ r = 5 $$ into the circle's equation:

$$
   (-3)^2 + y^2 = 5^2.
   $$

6. Compute:
   $$
   9 + y^2 = 25.
   $$

7. Solve for $$ y^2 $$:
   $$
   y^2 = 25 - 9 = 16.
   $$

8. Take the square root of both sides. Remember that both positive and negative roots are possible:
   $$
   y = \pm 4.
   $$

Thus, the solutions are:

- For the trigonometric equation:  
  $$
  \cos x = -\frac{3}{5} \quad \Longrightarrow \quad x = \arccos\left(-\frac{3}{5}\right) + 2\pi k \quad \text{or} \quad x = -\arccos\left(-\frac{3}{5}\right) + 2\pi k, \quad k \in \mathbb{Z}.
  $$

- For the circle equation with $$ x = -3 $$ and $$ r = 5 $$:  
  $$
  y = \pm 4.
  $$
The solutions are: - $$ \cos x = -\frac{3}{5} $$ with $$ x = \arccos\left(-\frac{3}{5}\right) + 2\pi k $$ or $$ x = -\arccos\left(-\frac{3}{5}\right) + 2\pi k $$ for any integer $$ k $$. - $$ y = 4 $$ or $$ y = -4 $$.

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