find the exact zeros of the function in the interval [0, 2π)
2 cot squared x + 3 csc x=0
We know that
1 + cot^2 Ø = csc^2 Ø
so our equation is
2(csc^ x - 1) + 3csc x = 0
2 csc^2 x + 3csc x - 2 = 0
(2cscx -1)(cscx + 2) = 0
cscx = 1/2 or cscx = -2
or
sinx = 2 (not possible) or sinx = -1/2
so x must be in III or IV
x = 7π/6 or 11π/6
To find the exact zeros of the given function, 2 cot^2(x) + 3 csc(x) = 0, within the interval [0, 2π), we will first simplify the equation by rewriting cot^2(x) and csc(x) in terms of sine and cosine.
First, we know that cot^2(x) can be rewritten as (cos(x) / sin(x))^2, and csc(x) is equal to 1 / sin(x). So our equation becomes:
2(cos^2(x) / sin^2(x)) + 3 / sin(x) = 0
Let's manipulate the equation further by multiplying through by sin^2(x) to clear the denominators:
2cos^2(x) + 3sin(x) = 0
Now, let's use trigonometric identities to rewrite cos^2(x) in terms of sin^2(x):
cos^2(x) = 1 - sin^2(x)
Substituting this back into our equation, we get:
2(1 - sin^2(x)) + 3sin(x) = 0
Multiplying out, we have:
2 - 2sin^2(x) + 3sin(x) = 0
Rearranging the terms, we get a quadratic equation:
2sin^2(x) - 3sin(x) - 2 = 0
Now, we can solve this quadratic equation for sin(x) using factoring, completing the square, or the quadratic formula.
Factoring the quadratic equation, we have:
(2sin(x) + 1)(sin(x) - 2) = 0
Setting each factor to zero and solving for sin(x), we get:
2sin(x) + 1 = 0 or sin(x) - 2 = 0
For the first factor, 2sin(x) + 1 = 0, subtracting 1 and dividing by 2:
2sin(x) = -1
sin(x) = -1/2
From the unit circle or trigonometric ratios, we know that sin(x) = -1/2 has solutions at π/6 and 5π/6 within the interval [0, 2π).
For the second factor, sin(x) - 2 = 0, adding 2:
sin(x) = 2
Since there are no solutions for sin(x) = 2 within the interval [0, 2π), this equation has no solutions within the given interval.
Therefore, the exact zeros of the function within the interval [0, 2π) are π/6 and 5π/6.