Question

Determine a general formula​ (or formulas) for the solution to the following equation.​ Then, determine the specific solutions​ (if any) on the interval [0,2 pi)

.
tangent theta= 0

Answer 1

Choice A: Only one equation will be sufficient for describing
`\theta` as in
`\tan \theta = 0`.

For
`\tan \theta` to be equal to
`0`,
`\theta = 0 + k \pi` where
`k` is a whole number (including zero or negative values.)

There are two specific solutions for
`\theta` on the interval
`[0,\, 2\pi)`:

• 
  `\theta= 0` (from
  `k = 0`,) and
• 
  `\theta = \pi` (from
  `k = 1`.)

Explanation:

In a right triangle, the tangent of an angle (other than the right angle) is equal to
`\displaystyle \frac{\text{Opposite}}{\text{Adjacent}}`.

In general, if
`\theta` corresponds to the point
`(x,\, y)` on the unit circle, then
`\displaystyle \tan \theta = (y)/(x)`.

This question asks that
`\tan \theta` shall be zero. In other words,
`\displaystyle (y)/(x) = 0`. The
`x`-coordinate is the denominator and can't be zero. The only possibility is that the
`y`-coordinate is zero. Graphically, there are two places on a unit circle where that could happen. (Wherefore the
`(2\pi)\, k`? On a unit circle, each turn is
`2\pi` radians. The angle would appear to land at the exact same position after one or more complete turns of

• 
  `y = 0` at
  `(1,\, 0)`, the intersection of the unit circle and the positive
  `x`-axis. At that point,
  `\theta = 0 + (2\pi)\, k`.
• 
  `y = 0` at
  `(-1,\,0)`,the intersection of the unit circle and the negative
  `x`-axis. At that point,
  `\theta = \pi + (2\pi)\, k`.

Note that these two values are separated by exactly
`\pi` radians. In other words, there is a zero every time
`\theta` goes through a half circle. After
`0`, there's

`\theta = 0 + k\pi`.

On the range
`0\le \theta < 2\pi`,

`0 \le 0 + k \pi <2\pi`.

`0 \le k\pi <2\pi`.

`0 \le k < 2`.

In other words,
`k = 0` or
`k = 1`. That corresponds to
`\theta = 0` and
`\theta = \pi`.