Question

How do you solve `Cos (pi/6 + x) + sin (pi/3 +x) = 0`?

Answer 1

`pi/2, (3pi)/2` for `(0, 2pi)`

Explanation:

Property of complementary arcs-->
`sin (pi/3 + x) = cos (pi/2 - (pi/3 + x)) = cos (pi/6 - x)`
Use trig identity:
`cos a + cos b = 2 cos ((a + b)/2).cos ((a - b)/2)`
The equation becomes:
`S = cos (pi/6 + x) + cos (pi/6 - x) = 0`
`a = (pi/6 + x)` , and `b = (pi/6 - x)`
`(a + b)/2 = pi/6`, and `(a - b)/2 = x`
Finally,
`S = 2cos (pi/6)cos x = 0`
cos x = 0 --> `x = pi/2` and `x = (3pi)/2`
Answers for `(0, 2pi)`:
`pi/2 ; (3pi)/2`
For general answers, add `2kpi`
Check with `x = (pi/2)`:
`cos (pi/6 + x) = cos (pi/6 + pi/2) = cos ((2pi)/3) = - 1/2`
`sin (pi/3 + x) = sin (pi/3 + pi/2) = sin ((5pi)/6) = 1/2`
`S = -1/2 + 1/2 = 0`. OK