How do you graph to solve the equation on the interval $[- 2 π, 2 π]$ $[-2pi,2pi]$ for $tan x = 1$ $tanx=1$ ?

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How do you graph to solve the equation on the interval $[- 2 π, 2 π]$ $[-2pi,2pi]$ for $tan x = 1$ $tanx=1$ ?

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Answer 1 · 2017-08-01T01:24:51

( $- \frac{7 π}{4}, - \frac{3 π}{4}, \frac{π}{4}, \frac{5 π}{4}$ $-(7pi)/4,-(3pi)/4,pi/4,(5pi)/4$ )

Explanation:

Solving for x:
$tan x = 1$ $tanx=1$
$x = arctan 1$ $x=arctan1$
When the tangent of an angle is 1, we know that the lengths of the non hypotenuse sides are equal and have the same sign. So: Please excuse the MS Paint
We know that the values of angles x and y are $\frac{π}{4}$ $pi/4$ , and we need to add those values to known values on the interval. The solutions will be ( $- 2 π + \frac{π}{4}$ $-2pi+pi/4$ , $- π + \frac{π}{4}$ $-pi+pi/4$ , $\frac{π}{4}$ $pi/4$ , $π + \frac{π}{4}$ $pi+pi/4$ ), which simplifies to:

( $- \frac{7 π}{4}, - \frac{3 π}{4}, \frac{π}{4}, \frac{5 π}{4}$ $-(7pi)/4,-(3pi)/4,pi/4,(5pi)/4$ )

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How do you graph to solve the equation on the interval $[- 2 π, 2 π]$ $[-2pi,2pi]$ for $tan x = 1$ $tanx=1$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you graph to solve the equation on the interval [−2π,2π][-2pi,2pi] for tanx=1tanx=1?

1 Answer

Explanation:

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Impact of this question

How do you graph to solve the equation on the interval $[- 2 π, 2 π]$ $[-2pi,2pi]$ for $tan x = 1$ $tanx=1$ ?