The number of solutions of the equation cot ⁡ x = cot ⁡ x + cosec x in 0,10 π is/are

Step-by-step solution

If x lies in 1 s t or 3 r d quadrant then cot ⁡ x = cot ⁡ x , thus equation becomes cot ⁡ x = cot ⁡ x + 1 sin ⁡ x ⇒ 1 sin ⁡ x = 0 , not possible If x lies in 2 n d or 4 t h quadrant, then cot ⁡ x = - cot ⁡ x , thus equation becomes - cot ⁡ x = cot ⁡ x + 1 sin ⁡ x ⇒ - 2 cot ⁡ x = 1 sin ⁡ x ⇒ - 2 cos ⁡ x sin ⁡ x = sin ⁡ x ⇒ sin ⁡ x 1 + 2 cos ⁡ x = 0 ⇒ cos ⁡ x = - 1 2 ∵ sin ⁡ x ≠ 0 ⇒ x = 2 π 3 is the only solution (In 2 n d quad.) In 0,2 π there is only one solution In 0,10 π , we get, five solutions