Question 1056110
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{{{cot("135°"-2y)}}}{{{""=""}}}{{{-0.5}}}

Cotangent is the reciprocal of the tangent:

{{{1/tan("135°"-2y)}}}{{{""=""}}}{{{-0.5}}}

Multiply both sides by tan(135°-2y):
 
{{{1}}}{{{""=""}}}{{{-0.5tan("135°"-2y)}}}

Divide both sides by -0.5:

{{{1/(-0.5)}}}{{{""=""}}}{{{tan("135°"-2y)}}}

Simplify left side:

{{{-2}}}{{{""=""}}}{{{tan("135°"-2y)}}}

Use formula for tan(A-B):

{{{-2}}}{{{""=""}}}{{{(tan("135°")-tan(2y))/(1+tan("135°")tan(2y))}}}

Use the fact that tan(135°) = -1, substituting:

{{{-2}}}{{{""=""}}}{{{((-1)-tan(2y))/(1+(-1)tan(2y))}}}

Simplify:

{{{-2}}}{{{""=""}}}{{{(-1-tan(2y))/(1-tan(2y))}}}

Multiply both sides by 1-tan(2y):

{{{-2(1-tan(2y)^"")}}}{{{""=""}}}{{{-1-tan(2y)}}}

Distribute on the left:

{{{-2+2tan(2y)}}}{{{""=""}}}{{{-1-tan(2y)}}}

Add +2 to both sides:

{{{2tan(2y)}}}{{{""=""}}}{{{1-tan(2y)}}}

Add + tan(2y) to both sides:

{{{3tan(2y)}}}{{{""=""}}}{{{1}}}

Divide both sides by 3:

{{{tan(2y)}}}{{{""=""}}}{{{1/3}}}

Use formula for tan(2&#952;):

{{{2tan^""(y)/(1-tan^2(y))}}}{{{""=""}}}{{{1/3}}}

Cross-multiply:

{{{6tan^""(y)}}}{{{""=""}}}{{{1-tan^2(y)}}}

Get 0 on the right:

{{{tan^2(y)+6tan^""(y)-1}}}{{{""=""}}}{{{0}}}

Use quadratic formula:

{{{tan(y) = (-6 +- sqrt( 6^2-4*1*(-1) ))/(2*1) }}}

Simplify:

{{{tan(y) = (-6 +- sqrt(36+4))/2 }}}

{{{tan(y) = (-6 +- sqrt(40))/2 }}}

{{{tan(y) = (-6 +- sqrt(4*10))/2 }}}

{{{tan(y) = (-6 +- 2sqrt(10))/2 }}}

{{{tan(y) = (2(-3 +- sqrt(10)))/2 }}}

{{{tan(y) = (cross(2)(-3 +- sqrt(10)))/cross(2) }}}

{{{tan(y) = -3 +- sqrt(10) }}}

Using the + :

{{{tan(y) = -3 + sqrt(10) }}}

{{{tan(y) = 0.1622776602}}}

{{{y = "9.217474413°"}}}  <-- 1st quadrant solution

Add 180° to get the 3rd quadrant solution:

{{{y = "189.217474413°"}}}  <-- 3rd quadrant solution

Using the - :

{{{tan(y) = -3 - sqrt(10) }}}

{{{tan(y) = -6.1622776602}}}

{{{y = "-80.78252559°"}}}  <-- 4th quadrant negative solution

which calculator always gives but which we cannot accept here
since it's not between 0° and 360°.

So we add 360° to get an equivalent positive 4th quadrant 
solution which is between 0° and 360:

{{{y = "279.217474413°"}}}  <-- 4th quadrant solution

Subtract 180° to get the 2nd quadrant solution:

{{{y = "99.217474413°"}}}  <-- 2nd quadrant solution

So there are 4 solutions between 0° and 360°.

Edwin</pre></b></font>