## moneybird Group Title What is the ordered pair of positive intgers (a,b) for which a/b is a reduced fraction and $x = \frac{a \pi}{b}$ is the least positive solution of the equation: $(2 \cos (8x) - 1)(2 \cos (4x) - 1)(2 \cos (2x) - 1)(2 \cos (x) - 1) = 1$? 2 years ago 2 years ago

1. agdgdgdgwngo Group Title

0?

2. moneybird Group Title

i don't know the solution

3. agdgdgdgwngo Group Title

$$2 \pi$$ ?

4. moneybird Group Title

how do you get that?

5. Mertsj Group Title

What course are you taking moneybird that this problem came from?

6. moneybird Group Title

7. Mertsj Group Title

ty

8. Mertsj Group Title

This equation will be true if each factor = 1. Each factor will equal 1 if x = 0. So a/b, which must be 2 positive integers is 2/1

9. moneybird Group Title

how about 1 factor is 4, and 1 factor is 1/4

10. Mertsj Group Title

That could not be true. The largest value of the cos is 1

11. Mertsj Group Title

So the largest value of 2cos x is 2

The maximum of every parenthetical is 1, the minimum is -2. You would need multiplicative inverses and even factors of positive for this to work

13. Mertsj Group Title

Regardless of the value of x

14. moneybird Group Title