![\bf cot(x)cos(x)-cot(x)=0\implies cot(x)[cos(x)-1]=0 \\\\[-0.35em] ~\dotfill\\\\ cot(x)=0\implies \cfrac{cos(x)}{sin(x)}=0\implies cos(x)=0\\\\\\ x=cos^{-1}(0)\implies \boxed{x= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}} \\\\[-0.35em] ~\dotfill\\\\ cos(x)-1=0\implies cos(x)=1\implies x=cos^{-1}(1)\implies \boxed{x=0}](/tpl/images/0400/1977/b186a.png)
cotxcosx - cotx = 0
0
pi/2
Mathematics, 03.12.2019 01:31 nia3378
Solve for x for 0 ≤ x < 2 π .
cotxcosx - cotx = 0
0
pi/2
pi
3pi/2
2pi
Answers: 2
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You know the right answer?
Solve for x for 0 ≤ x < 2 π .
cotxcosx - cotx = 0
0
pi/2
cotxcosx - cotx = 0
0
pi/2
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