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

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Q. The number of solutions of the equation $tan xsin⁡x-1=tan⁡x-sin⁡x,\forall x\in \left[0,2 \pi \right]$ is equal to.

NTA AbhyasNTA Abhyas 2020

Solution:

$tan xsin⁡x-tan⁡x+\left(sin ⁡ x - 1\right)=0$
$\Rightarrow \left(tan x + 1\right)\left(sin ⁡ x - 1\right)=0$
$\Rightarrow tan x = - 1$ or $sin x = 1$ (Not possible as $x\neq \frac{\pi }{2}$ )
Hence, for $tanx=-1$ there are $2$ solutions in $\left[0 , 2 \pi \right]$