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.