Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given,

cosec x= -\(\sqrt2\)

We know that sin x, and cosec x have negative values in the 3^rd and 4^th quadrant. 

While giving a solution, we always try to take the least value of y 

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e., negative angle) 

- √2 = -cosec \((\frac{π}4)\) = cosec \((-\frac{π}4)\) { ∵ sin - θ = - sin θ}

∴ cosec x = cosec\((-\frac{π}4)\)

⇒ sim x = sin \((\frac{-π}4)\)

If sin x = sin y ,then x = nπ + (– 1)ⁿy , where n ∈ Z. 

For above equation y = \(-\frac{π}4\)

∴ x = nπ + (-1)ⁿ \((-\frac{π}4)\) ,where n ϵ Z

Or x = nπ + (-1)ⁿ⁺¹ \((\frac{π}4)\),where n ϵ Z 

Thus, 

x gives the required general solution for given trigonometric equation.