Question

What is the general solution of `cos(x)=0` if `3^(cos(x))=1` with `-180^@ <= x <= 360^@`?

3 cos x =1
where -180º ≤ x ≤ 360º
find general solution of
cos x = 0 `a^2 + b^2 = c^2`

Answer 1

`x=-90^@`, `90^@`, and `270^@`

Explanation:

The rules of exponents tells us that

`3^0=1`

You are given that

`3^(cos(x))=1`

This means that

`3^0=3^(cos(x))`

This last equation can only be true if the exponents are equal.

`cos(x)=0`

If you look at the unit circle, and understand that `cos(x)` corresponds to the `x`-coordinates, then you will see that `cos(x)=0` always and only on the vertical, `y`-axis.

For the negative angles, you move clockwise from `x=0^@`. So when `-180^@ <= x <= 0^@`, you only hit the `y` axis once at `x=-90^@`.

For the positive angles, you move counter clockwise from `x=0^@`. So when `0^@ <= x <= 360^@`, you hit the `y`-axis once when you hit `90^@` and a second time when you hit it again at `270^@`