The number of solutions of the equation | cot x | = cot x + 1 sin x in the interval [ 0 ,   2 π ] is

The number of solutions of the equation |cotx|=cotx+1sinx in the interval [0, 2π] is

Solution

If cotx>0⇒1sinx=0 (Not possible)

If cotx<0⇒2cotx+1sinx=0

⇒2cosx=-1

⇒x=2π3 or 4π3 (reject as cotx<0)

So, number of solutions is one.

Asked in: JEE Main 2021 (18 Mar Shift 1)

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