Find the general solutions of the following equations :


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,



We know that sin x, and cosec x have negative values in the 3rd and 4th 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 (π/4) = cosec (-π/4) { sin -θ = -sin θ }




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


For above equation y =


x = nπ + (-1)n,where n ϵ Z


Or x = nπ + (-1)n+1 ,where n ϵ Z


Thus, x gives the required general solution for given trigonometric equation.


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