We are given two trigonometric equations:

$1 - \cos(5a) + \cos(5b) = 0 \quad \text{(1)}$ $1 - \cos(7a) + \cos(7b) = 0 \quad \text{(2)}$

Step 1: Rearrange each equation

Equation (1):

Rearranging for $\cos(5a)$:

$1 - \cos(5a) + \cos(5b) = 0$ $\cos(5a) = 1 + \cos(5b) \quad \text{(3)}$

Equation (2):

Similarly, rearrange equation (2):

$1 - \cos(7a) + \cos(7b) = 0$ $\cos(7a) = 1 + \cos(7b) \quad \text{(4)}$

Step 2: Apply cosine bounds

The cosine function has a range of $[-1, 1]$, meaning for any real value of $a$ and $b$, the values of $\cos(5a)$ and $\cos(7a)$ must lie within this range. This helps us consider special cases when analyzing further.

Step 3: Try specific values

We can test specific values of $a$ and $b$ to see if they satisfy both equations simultaneously. A logical choice is $a = b$, as it might lead to a simpler solution.

If $a = b$, then from equation (3):

$\cos(5a) = 1 + \cos(5a)$

This simplifies to:

$\cos(5a) = 1$

Thus, $5a = 2n\pi$, meaning:

$a = \frac{2n\pi}{5} \quad \text{for integer } n.$

Now apply the same logic to equation (4):

$\cos(7a) = 1 + \cos(7a)$

This simplifies to:

$\cos(7a) = 1$

Thus, $7a = 2m\pi$, meaning:

$a = \frac{2m\pi}{7} \quad \text{for integer } m.$

Step 4: Solve for $a$ and $b$

For $a = b$, both $a$ and $b$ must satisfy:

$\frac{2n\pi}{5} = \frac{2m\pi}{7}$

This implies $5m = 7n$, so $m = 7k$ and $n = 5k$ for some integer $k$. Therefore:

$a = b = \frac{2k\pi}{35}$

Conclusion:

The general solutions for $a$ and $b$ are:

$a = b = \frac{2k\pi}{35} \quad \text{for integer } k.$

Would you like further details or clarifications on any step?

Related Questions:

1. What is the solution if $a \neq b$?
2. How does the cosine function behave for large values of $a$ and $b$?
3. Can you solve the equations graphically?
4. What happens if we use sine functions instead of cosine?
5. How does the solution change if $a$ or $b$ are complex?

Tip: Cosine values are periodic, so checking the behavior at specific intervals like multiples of $\pi$ often simplifies the problem.