Cscx=cot^2x-1... {0<x<2pi}?

Question

cscx=cot^2x-1... {0<x<2pi}

thanks!

Answer 1

Well we need to use a pythagorean identity to help us get one trig funciton.

We know that cot^2x = csc^2x - 1, substituting, we get:

cscx = csc^2x - 1 - 1

cscx = csc^2x - 2  we know have only cscx so and one of them is squared, so let&#039;s use quadratics:

csc^2x - cscx - 2 = 0  factor

(cscx - 2)(cscx + 1) = 0

set each term to zero:

cscx = 2  cscx = -1  csc is the reciprocal of sin

sinx = 1/2  sinx = -1  use your unit circle to find the solutions

x = pi/6, 5pi/6, 3pi/2

voila