Question

How do you solve `sin2x+sin4x=cos2x+cos4x` for x in the interval [0,2pi)?

Answer 1

`pi/12 and (5pi)/12`

Explanation:

Apply the sum- to-product identities:
`sin a + sin b = 2sin ((a + b)/2).cos ((a -b)/2)`.
In this case:
`sin 2x + sin 4x = 2sin ((2x + 4x)/2).cos ((4x - 2x)/2) =`
`= 2sin (3x).cos x` (1)
`cos 2x + cos 4x = 2cos ((2x + 4x)/2).cos ((4x - 2x)/2) =`
`= 2cos (3x).cos x.`(2)
We have : (1) = (2)
2sin (3x).cos x = 2cos (3x).cos x
After simplification:
sin (3x) = cos (3x)
Divide both sides by cos 3x
`tan 3x = 1 = tan (pi/4)` -->
Trig table and unit circle give 2 solutions:
a. `3x = pi/4` --> `x = pi/12.`and
b. `3x = pi/4 + pi = 5pi/4`--> `x = (5pi)/12`
Check.
`x = pi/12 = 180/12 = 15^@` --> `2x = 30^@` --> `4x = 60^@`
`sin 30 + sin 60 = 1/2 + sqrt3/2`
`cos 30 + cos 60 = sqrt3/2 + 1/2`. Proved