Question 1206427
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Answers:
{{{x = 0}}}, {{{x = 3pi/4}}}, {{{x = pi}}}, {{{x = 7pi/4}}}


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Work Shown


{{{tan^2(x) + tan(x) = 0}}}


{{{tan(x)(tan(x) + 1) = 0}}}


{{{tan(x)=0}}} or {{{tan(x) + 1 = 0}}}


{{{tan(x)=0}}} or {{{tan(x) = -1}}}


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{{{tan(x) = 0}}}


{{{sin(x)/cos(x) = 0}}}


{{{sin(x) = cos(x)*0}}}


{{{sin(x) = 0}}}


This leads to {{{x = 0}}} and {{{x = pi}}} when considering the interval [0, 2pi)
Refer to the unit circle. These angles occur when the y coordinate is 0. 


We exclude x = 2pi since [0, 2pi) means {{{0 <= x < 2pi}}}


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{{{tan(x) = -1}}}


{{{sin(x)/cos(x) = -1}}}


{{{sin(x) = -cos(x)}}}


Look on the unit circle for terminal points where x = -y or y = -x. 
In other words, look for points on the unit circle that have the same coordinates just one is negative.


Tangent is negative in quadrants II and IV


The solution in quadrant II is {{{x = 3pi/4}}} and the solution in quadrant IV is {{{7pi/4}}} (they are separated by a gap of pi radians aka 180 degrees).
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