The number of solutions of the equation cot xcos⁡x-1=cot⁡x-cos⁡x, ∀ x∈ [0 , 2 π ] is equal to

Question

The number of solutions of the equation cot xcos⁡x-1=cot⁡x-cos⁡x, ∀ x∈ [0 , 2 π ] is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

cot x ⋅ cos x - cot x + (cos x - 1) = 0 ⇒ (cot x + 1)(cos ⁡ x - 1)=0 ⇒ cot x=-1 or cos x=1 (Not possible as sin x≠ 0 ) Hence, 2 solutions are possible in [0 , 2 π ]

Q. The number of solutions of the equation $co t x cos x - 1 = co t x - cos x, \forall x \in [0, 2 π]$ is equal to

$co t x \cdot cos x - co t x + (cos x - 1) = 0$
$\Rightarrow (co t x + 1) (cos x - 1) = 0$
$\Rightarrow co t x = - 1$ or $cos x = 1$ (Not possible as $s in x \neq = 0$ )
Hence, $2$ solutions are possible in $[0, 2 π]$

Q. The number of solutions of the equation cotxcos⁡x−1=cot⁡x−cos⁡x,∀x∈[0,2π] is equal to

cotx⋅cosx−cotx+(cosx−1)=0 ⇒(cotx+1)(cos⁡x−1)=0 ⇒cotx=−1 or cosx=1 (Not possible as sinx=0 ) Hence, 2 solutions are possible in [0,2π]

Q. The number of solutions of the equation $co t x cos x - 1 = co t x - cos x, \forall x \in [0, 2 π]$ is equal to

$co t x \cdot cos x - co t x + (cos x - 1) = 0$
$\Rightarrow (co t x + 1) (cos x - 1) = 0$
$\Rightarrow co t x = - 1$ or $cos x = 1$ (Not possible as $s in x \neq = 0$ )
Hence, $2$ solutions are possible in $[0, 2 π]$