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Hi

Rewrite cot^2(x) in terms of csc(x) using the pythagorean identity, then use the quadratic formula to solve it.

2cot^2(x) + 3csc(x) = 0

2[csc^2(x) - 1] + 3csc(x) = 0

2csc^2(x) - 2 + 3csc(x) = 0

2csc^2(x) + 3csc(x) - 2 = 0

csc(x) = {-3 ± √[3^2 - 4(2)(-2)]}/[2(2)]

csc(x) = (-3 ± 5)/4 = 1/2 or -2

csc(x) = 1/2 or csc(x) = -2

<no solution> or x = arccsc(-2) = -π/6

So the answer is x = -π/6.

I hope this helps!

Keep in mind that cot²x = csc²x - 1.

You can replace it so that they will all be cscx.

2cot²x + 3cscx = 0

2(csc²x - 1) + 3cscx = 0

2csc²x - 2 + 3cscx = 0

2csc²x + 3cscx - 2 = 0

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

2cscx - 1 = 0

2cscx = 1

cscx = 1/2

x = non real

cscx + 2 = 0

cscx = -2

x = -π/6

x = -π/6 is the only real result.

Hope this helps!

2cot²x + 3cscx = 0

[ 2cos^2(x) / sin^2(x) ] + [ 3 / sin(x) ] = 0

[ 2cos^2(x) / sin^2(x) ] = [- 3 / sin(x) ]

[ 2cos^2(x) / sin(x) ] = [- 3 ]

[ 2cos^2(x) ] = [- 3 sin(x) ]

2cos^2(x) + 3 sin(x) = 0

2[ 1 - sin^2(x) ] + 3 sin(x) = 0

2 - 2sin^2(x) + 3 sin(x) = 0

2sin^2(x) - 3 sin(x) - 2 = 0 ======> calling sin(x) = a

2a^2 - 3a - 2 = 0

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

2a + 1 = 0 ====> 2a = -1 ====> a = -1/2 =====>

sin(x) = -1/2 ===> sin^-1 ( sin(x) ) = sin^-1(-1/2) ====> x = [ π + π/6 = 7π/6 ] & [ 2π - π/6 = 11π/6]

a = 2 ====> sin(x) = 2 ===> sin^-1 ( sin(x) ) = sin^-1(2) ====> x = not defined because it is greater than one