How do you solve $cotx=cosx$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-08-19T03:13:11

${pi/2, (3pi)/2}$

$cosx/sinx - cosx = 0$

$(cosxsinx - cosx)/sinx = 0$

$cosxsinx - cosx = 0$

$cosx(sinx - 1) = 0$

$cosx = 0 and sinx = 1$

$x = pi/2, (3pi)/2 and pi/2$

Note that this equation has restrictions of $x != pi, 0$ , since it would render the equation undefined.

Hopefully this helps!

How do you solve cotx=cosxcotx=cosx in the interval 0<=x<=2pi0<=x<=2pi?