How do you solve $cot^2x-cscx=1$ and find all solutions in the interval $0<=x<360$ ?

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Answer 1 · 2016-10-17T13:20:23

$cos^2x/sin^2x - 1/sinx = 1$

$cos^2x/sin^2x -sinx/sin^2x - sin^2x/sin^2x = 0$

$cos^2x -sinx - sin^2x = 0$

Apply the identity $cos^2x = 1 - sin^2x$ .

$1 - sin^2x - sinx - sin^2x = 0$

$-2sin^2x - sinx + 1 = 0$

$-2sin^2x - 2sinx + sinx + 1 = 0$

$-2sinx(sinx + 1) + 1(sinx + 1) =0$

$(-2sinx + 1)(sinx + 1) = 0$

$sinx =-1/2 and sinx = -1$

$x = 210˚, 330˚, 270˚$

Hopefully this helps!

How do you solve cot^2x-cscx=1cot^2x-cscx=1 and find all solutions in the interval 0<=x<3600<=x<360?