find the exact solutions over the interval
tanx = 1 all real x
you're kidding, right?
tan pi/4 = 1
x = k*pi + pi/4 for any integer k
No, but thank for the help
To find the exact solutions of the equation tan(x) = 1 over the interval of all real x, we need to determine the values of x that satisfy this equation.
The equation tan(x) = 1 can be rewritten as x = arctan(1). However, the arctan function has multiple solutions, since it is periodic with a period of π. Therefore, we need to consider the general solution.
The general solution for the equation tan(x) = 1 is x = arctan(1) + nπ, where n is an integer.
To find the exact solutions over the interval of all real x, we need to substitute different values for n and solve for x.
Let's substitute n = 0 first:
x = arctan(1) + 0π
x = arctan(1)
The principal value of arctan(1) is π/4. Therefore, the first solution is x = π/4.
Now let's substitute n = 1:
x = arctan(1) + 1π
x = arctan(1) + π
The principal value of arctan(1) is π/4. Therefore, the second solution is x = π/4 + π = 5π/4.
We can find additional solutions by substituting different values for n.
If we substitute n = -1:
x = arctan(1) - 1π
x = arctan(1) - π
The principal value of arctan(1) is π/4. Therefore, the third solution is x = π/4 - π = -3π/4.
Similarly, we can find more solutions by substituting different values for n.
Therefore, the exact solutions over the interval of all real x are:
x = π/4, 5π/4, -3π/4, 9π/4, -7π/4, ...
Note that the pattern continues with an integer multiple of π added or subtracted from π/4.