How do you solve $cot^2x-cscx=1$ ?

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Answer 1 · 2017-01-03T15:32:51

$x = pi/6 + 2pin, (5pi)/6 + 2pin and (3pi)/2 + 2pin$

Use the pythagorean identity $cot^2theta + 1 = csc^2theta$ . This means that $cot^2theta = csc^2theta - 1$ .

$csc^2x - 1 - cscx = 1$

$csc^2x - cscx - 2 = 0$

Let $t = cscx$ .

$t^2 - t - 2 = 0$

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

$t = 2 and -1$

$cscx = 2 and cscx = -1$

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

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

$x = pi/6 + 2pin, (5pi)/6 + 2pin and (3pi)/2 + 2pin$

Hopefully this helps!

How do you solve cot^2x-cscx=1cot^2x-cscx=1?