How do I solve the equation $cot (\frac{5 π}{12} - x) + 1 = 0$ $cot({5pi}/12-x)+1=0$ ?

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Answer 1 · 2018-06-08T11:44:14

$x = {2pi}/3 + pi k quad$ integer $k$

$cot({5pi}/12 - x) + 1 = 0$

$cot({5pi}/12 - x) = - 1$

Cotangents are reciprocal slopes, so a cotangent of $-1$ is a tangent of $1/-1=-1$ too so $-45^circ$ or $135^circ$ and their coterminal friends.

$cot({5pi}/12 - x) = cot ( - pi/4)$

${5 pi}/12 - x = -pi/4 + pi k quad$ integer $k$

$x = {5pi}/12 + pi/4 + pi k quad$ integer $k$

(It's OK to flip the sign on $k$ , which still ranges over the integers.)

$x = {2pi}/3 + pi k quad$ integer $k$

Check: Let's just check $k=-1$ so $x=-pi/3$

$cot({5pi}/12 - (-pi/3)) + 1 = cot( {3pi}/4) + 1 = -1 + 1 = 0 quad sqrt$

How do I solve the equation cot({5pi}/12-x)+1=0cot(5π12−x)+1=0cot({5pi}/12-x)+1=0?