Easy trig. (#3) (1 Viewer)

[sinophile]

Solve for x: cot(x+60)=1, 0<=x<=360.

My solution:
cot(x+60)=1 (60<=x+60<=420)
1/tan(x+60)=1
tan(x+60)=1
Related angle is tan^1(1)=45

Since cot(x+60) is positive, x+60 lies in 1st and 3rd quadrant.
Therefore x+60=45, 225
x= -15, 165

Textbook solution:
x=165, 345.

Why does the textbook solution not include x=-15, when x=-15 lies within the domain of x? Why, instead, does it suggest 345, which is an equivalent angle? Or am I missing something here? Thanks!

 

[Trebla]

Solve for x: cot(x+60)=1, 0<=x<=360.

My solution:
cot(x+60)=1 (60<=x+60<=420)
1/tan(x+60)=1
tan(x+60)=1
Related angle is tan^1(1)=45

Since cot(x+60) is positive, x+60 lies in 1st and 3rd quadrant.
Therefore x+60=45, 225
x= -15, 165

Textbook solution:
x=165, 345.

Why does the textbook solution not include x=-15, when x=-15 lies within the domain of x? Why, instead, does it suggest 345, which is an equivalent angle? Or am I missing something here? Thanks!

If:
0 < x < 360, then
60 < x + 60< 420

 

[Timothy.Siu]

Solve for x: cot(x+60)=1, 0<=x<=360.

My solution:
cot(x+60)=1 (60<=x+60<=420)
1/tan(x+60)=1
tan(x+60)=1
Related angle is tan^1(1)=45

Since cot(x+60) is positive, x+60 lies in 1st and 3rd quadrant.
Therefore x+60=45, 225
x= -15, 165

Textbook solution:
x=165, 345.

Why does the textbook solution not include x=-15, when x=-15 lies within the domain of x? Why, instead, does it suggest 345, which is an equivalent angle? Or am I missing something here? Thanks!

clearly -15 doesn't lie in the domain of x which is 0<=x<=360.

 

[youngminii]

0 <= x <= 360
This means that the FINAL solution for x MUST be between 0 and 360.
Your answers are -15 and 165. -15 is NOT in between 0 and 360. What is the same as cot(-15) but inside 0 and 360? cot(345). And that is why x = 345 instead of -15.

 

[sinophile]

Okay, thank you all for your replies! Im sorry I have annoyed you all so many times.