Solve the following equation for x, where 0 </ x 2(pi).
tan^2x=1
Convert --→ 4.2 g = ______ kg.
Umm...what?
Solve the following equation for x, where 0 ≤ x < π
tan²x=1
Between 0 and π, tan π/4=1,
so what is x?
To solve the equation tan^2(x) = 1, where 0 ≤ x ≤ 2π, we can use the knowledge that the tangent function has a period of π. Therefore, we can find the values of x in the interval [0, 2π) where tan^2(x) equals 1.
First, let's consider the equation tan^2(x) = 1. Since tan(x) is the square root of tan^2(x), we can write a new equation as tan(x) = ±√1. Taking the square root of 1 results in ±1. So, we now have two separate equations:
tan(x) = 1
tan(x) = -1
Now, let's find the values of x for each equation separately.
1) tan(x) = 1:
To find the values of x where tan(x) equals 1, we can use the inverse tangent function, denoted as atan or arctan. Take the arctan of both sides of the equation:
atan(tan(x)) = atan(1)
This simplifies to:
x = atan(1)
Using a calculator, we find that atan(1) = π/4. So, one solution in the interval [0, 2π) is x = π/4.
2) tan(x) = -1:
To find the values of x where tan(x) equals -1, we again use the arctan function:
atan(tan(x)) = atan(-1)
This simplifies to:
x = atan(-1)
Using a calculator, we find that atan(-1) = -π/4. However, we need to find solutions in the interval [0, 2π), so we add 2π to account for the negative angle. Therefore, another solution in the given interval is x = -π/4 + 2π = 7π/4.
In summary, the solutions to the equation tan^2(x) = 1 in the interval [0, 2π) are x = π/4 and x = 7π/4.