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.