Question

How do you solve the equation `cos(x+pi/3)+cosx=0` on interval [0,2pi)?

Answer 1

`pi/3 and (4pi)/3`

Explanation:

Apply the trig identity:
`cos a + cos b = 2cos ((a + b)/2).cos ((a - b)/2)`
`cos (x + pi/3) + cos x = 2cos (x + pi/6)cos (pi/6) = 0`
Trig table --> cos pi/6 = sqrt3/2. The equation becomes:
`sqrt3(cos x + pi/6) = 0`
`cos (x + pi/6) = 0` --> 2 solutions:
a. `x + pi/6 = pi/2` -->`x = pi/2 - pi/6 = (2pi)/6 = pi/3`
b. `x + pi/6 = (3pi)/2` -->`x = (3pi)/2 - pi/6 = (8pi)/6 = (4pi)/3`
Answers for `0, 2pi)`
`pi/3 and (4pi)/3`
Check.
`x = (4pi)/3`--> `cos x = - cos pi/3 = -1/2.`
`cos (x + pi/3) = cos (5pi)/3 = cos pi/3 = 1/2`,
`cos x + cos (x + pi/3) = - 1/2 + 1/2 = 0.` OK