The number of solutions of the equation tan xsin⁡x-1=tan⁡x-sin⁡x,∀ x∈ [0,2 π ] is equal to.

Question

The number of solutions of the equation tan xsin⁡x-1=tan⁡x-sin⁡x,∀ x∈ [0,2 π ] is equal to. (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

tan xsinâ¡x-tanâ¡x+(sin â¡ x - 1)=0 ⇒ (tan x + 1)(sin â¡ x - 1)=0 ⇒ tan x = - 1 or sin x = 1 (Not possible as x≠ (π /2) ) Hence, for tanx=-1 there are 2 solutions in [0 , 2 π ]

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

$1$

$2$

$3$

$4$

$t an x s in x - t an x + (s in x - 1) = 0$
$\Rightarrow (t an x + 1) (s in x - 1) = 0$
$\Rightarrow t an x = - 1$ or $s in x = 1$ (Not possible as $x \neq = \frac{π}{2}$ )
Hence, for $t an x = - 1$ there are $2$ solutions in $[0, 2 π]$

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

1

2

3

4

tanxsin⁡x−tan⁡x+(sin⁡x−1)=0 ⇒(tanx+1)(sin⁡x−1)=0 ⇒tanx=−1 or sinx=1 (Not possible as x=2π​ ) Hence, for tanx=−1 there are 2 solutions in [0,2π]

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

$1$

$2$

$3$

$4$

$t an x s in x - t an x + (s in x - 1) = 0$
$\Rightarrow (t an x + 1) (s in x - 1) = 0$
$\Rightarrow t an x = - 1$ or $s in x = 1$ (Not possible as $x \neq = \frac{π}{2}$ )
Hence, for $t an x = - 1$ there are $2$ solutions in $[0, 2 π]$