Find all solutions in the interval [0, pi], 2sec^2x+tan^2x-3=0
To find all solutions in the interval [0, π] for the equation 2sec^2x + tan^2x - 3 = 0, we will first manipulate the equation to solve for sec^2x.
Let's start by moving the 3 to the right side of the equation:
2sec^2x + tan^2x = 3
Next, we can substitute sec^2x = 1 + tan^2x into the equation:
2(1 + tan^2x) + tan^2x = 3
Simplifying further:
2 + 2tan^2x + tan^2x = 3
3tan^2x + 2 = 3
3tan^2x = 1
tan^2x = 1/3
Now, let's solve for tan(x):
Taking the square root of both sides gives us:
tan x = ±√(1/3)
In the interval [0, π], the tangent function is positive in the first and third quadrants. Therefore, we need to find the angles where tan x is equal to √(1/3) and -√(1/3) in those quadrants.
To find the angles, we can use the inverse tangent function (also known as arctan or tan^(-1)).
For tan x = √(1/3), we have:
x1 = arctan(√(1/3))
For tan x = -√(1/3), we have:
x2 = arctan(-√(1/3))
Using a calculator in radian mode, we can evaluate the arctan(√(1/3)) and arctan(-√(1/3)) to get the approximate values:
x1 ≈ 0.61548
x2 ≈ -0.61548
Now, we need to check if these angles fall within the interval [0, π]. Since x1 is in the first quadrant, it satisfies the condition. However, x2 is in the fourth quadrant and is less than 0, which is outside the interval.
Therefore, the only solution in the interval [0, π] for the equation 2sec^2x + tan^2x - 3 = 0 is x ≈ 0.61548.