# What values of `theta` satisfy the equation `|2tan(theta)-1|=1 ` ?

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To solve this question, we must first remove the absolute value bars, thus allowing us to solve for each possible value. Recall, this means we can have to possible values for the expression inside the absolute value bars:

`2tantheta - 1 = +-1`

Let's solve for the first case:

`2tantheta-1 = 1`

Solve first for `tantheta`:

`2tantheta = 2`

`tantheta = 1`

Now, we need to find the `theta` for which `tantheta = 1`. You may recall this happens twice on the unit circle, represented by the following two angles:

`theta = pi/4` or `theta = 5/4 pi`

Like any other trigonometric function, tangent is periodic. However, unlike the other trigonometric functions, tangent has a period of `pi`, not `2pi`. Therefore, our full solution set is represented by the following relation, where `k_n` is an integer:

`theta = pi/4 + pik_1`

We can now solve for the second case, where the result is negative:

`2tantheta-1 = -1`

Let's solve for `tantheta` again:

`2tantheta = 0`

`tantheta = 0`

Now, we need to remember where `tantheta = 0`. Again, there are two places on the unit circle where we find this result:

`theta = 0` or `theta = pi`

Again, `tantheta` has a period of `pi`, so we can generalize the above solution to the following full set of solutions:

`theta = pik_2`

Now, we have the **full possible solution set** for the original equation:

`theta = pi/4 + pik_1` or `theta = pik_2`

We cannot easily combine these two solutions, so we must stop here.

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