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 can use the trigonometric identities relating sec(x) and tan(x) to sine(x) and cosine(x).

First, let's express sec^2x and tan^2x in terms of sine and cosine:

Recall that sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x). From these identities, we can write:

sec^2x = (1/cos(x))^2 = 1/cos^2(x)
tan^2x = (sin(x)/cos(x))^2 = sin^2(x)/cos^2(x)

Substituting these expressions into the equation, we get:

2/cos^2(x) + sin^2(x)/cos^2(x) - 3 = 0

To simplify further, let's find a common denominator. The common denominator for cos^2(x) and 1 is cos^2(x). For sin^2(x), we can write it as (1 - cos^2(x)) using the Pythagorean identity sin^2(x) + cos^2(x) = 1. So we have:

2/cos^2(x) + (1 - cos^2(x))/cos^2(x) - 3 = 0

Combining the terms over the common denominator gives us:

(2 + 1 - cos^2(x) - 3cos^2(x))/cos^2(x) = 0

Simplifying further, we get:

(3 - 4cos^2(x))/cos^2(x) = 0

Now, let's bring everything to one side of the equation:

3 - 4cos^2(x) = 0

Rearranging, we have:

4cos^2(x) = 3

Dividing both sides by 4, we get:

cos^2(x) = 3/4

Taking the square root of both sides, we have:

cos(x) = ± √(3/4)

Since we're looking for solutions in the interval [0, π], we need to consider the positive and negative values of √(3/4) separately.

1) For cos(x) = √(3/4):
In the interval [0, π], this solution occurs at x = π/6.

2) For cos(x) = -√(3/4):
In the interval [0, π], the cosine function is negative in the second and third quadrants. Using the symmetry of the cosine function around the y-axis, we can find the corresponding value for x in the second quadrant as follows:

cos(x) = cos(π - x)

So, if cos(x) = -√(3/4), then cos(π - x) = √(3/4). This means that x = π - π/6 = 5π/6 is another solution.

Therefore, the solutions in the interval [0, π] for the equation 2sec^2x + tan^2x - 3 = 0 are x = π/6 and x = 5π/6.