cotx - cosx = 1 - cotx. cosx માટે | vedclass

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Question

$\cot x-\cos x=1-\cot x \cos x$

$1-\cot x+\cos x-\cot x \cos x=0$

$(1-\cot x)(1+\cos x)=0$

$	herefore \cot x=1$ or $\cos x=-1$

$x=n \pi+\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\cot x-\cos x=1-\cot x \cos x$

$1-\cot x+\cos x-\cot x \cos x=0$

$(1-\cot x)(1+\cos x)=0$

$	herefore \cot x=1$ or $\cos x=-1$

$x=n \pi+\frac{\pi}{4}$ or $x=(2 n+1) \pi$

$	herefore$ Solution set $=x: x=2 n \pi+\pi, n \in I \cup x: x=2 n \pi+\frac{\pi}{4}, n \in I$

$cotx - cosx = 1 - cotx. cosx$ માટે $ x \in \left[ {0,2\pi } \right]$ ............ કિમતો મળે

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