Property of complementary arcs-->
sin (pi/3 + x) = cos (pi/2 - (pi/3 + x)) = cos (pi/6 - x)sin(π3+x)=cos(π2−(π3+x))=cos(π6−x)
Use trig identity:
cos a + cos b = 2 cos ((a + b)/2).cos ((a - b)/2)cosa+cosb=2cos(a+b2).cos(a−b2)
The equation becomes:
S = cos (pi/6 + x) + cos (pi/6 - x) = 0S=cos(π6+x)+cos(π6−x)=0
a = (pi/6 + x)a=(π6+x) , and b = (pi/6 - x)b=(π6−x)
(a + b)/2 = pi/6a+b2=π6, and (a - b)/2 = xa−b2=x
Finally,
S = 2cos (pi/6)cos x = 0S=2cos(π6)cosx=0
cos x = 0 --> x = pi/2x=π2 and x = (3pi)/2x=3π2
Answers for (0, 2pi)(0,2π):
pi/2 ; (3pi)/2π2;3π2
For general answers, add 2kpi2kπ
Check with x = (pi/2)x=(π2):
cos (pi/6 + x) = cos (pi/6 + pi/2) = cos ((2pi)/3) = - 1/2cos(π6+x)=cos(π6+π2)=cos(2π3)=−12
sin (pi/3 + x) = sin (pi/3 + pi/2) = sin ((5pi)/6) = 1/2sin(π3+x)=sin(π3+π2)=sin(5π6)=12
S = -1/2 + 1/2 = 0S=−12+12=0. OK
