may i have some help

Question

find the values of t in the interveral [0,2π] that satisify 2cos t +1=0

Answer 1

2cos t +1=0

 

cos t = -1/2

 

cosine is negative in Quadrants II and III

because x is negative there and x := r cos t.

(":=" means "is defined as")

 

Recall that in the 30°-60°-90° Special Triangle 1/2 = sin 30° = cos 60°.

 

Quadrants II and III angles with 60° = pi/3 reference angles are pi ± pi/3.

 

So the answers are 2pi/3 and 4pi/3.