Solve 2tan^2x + 1 = 0
Interval : [0, 2 pi]
To solve the equation 2tan^2x + 1 = 0 within the interval [0, 2π], we need to isolate the variable x. Here's how to do it step by step:
Step 1: Subtract 1 from both sides of the equation:
2tan^2x = -1
Step 2: Divide both sides of the equation by 2:
tan^2x = -1/2
Step 3: Take the square root of both sides of the equation:
tanx = ±√(-1/2)
Step 4: Find the values of x within the given interval where tanx is equal to ±√(-1/2).
First, let's find the angles where tanx = √(-1/2):
Using the inverse tangent function, we have:
x = arctan(√(-1/2))
Now let's find the angles where tanx = -√(-1/2):
Using the inverse tangent function, we have:
x = arctan(-√(-1/2))
Step 5: Calculate the values of x for both cases.
For tanx = √(-1/2):
We can use a calculator to find the principal value of x within the interval [0, 2π] where tanx = √(-1/2).
Using the calculator, we find:
x ≈ 0.588, 2.153
For tanx = -√(-1/2):
We can also use a calculator to find the principal value of x within the interval [0, 2π] where tanx = -√(-1/2).
Using the calculator, we find:
x ≈ 3.730, 5.296
So, the solutions for the equation 2tan^2x + 1 = 0 within the interval [0, 2π] are approximately:
x ≈ 0.588, 2.153, 3.730, 5.296.