Answer:
x = {nπ -π/4, (4nπ -π)/16}
Step-by-step explanation:
It can be helpful to make use of the identities for angle sums and differences to rewrite the sum:
cos(3x) +sin(5x) = cos(4x -x) +sin(4x +x)
= cos(4x)cos(x) +sin(4x)sin(x) +sin(4x)cos(x) +cos(4x)sin(x)
= sin(x)(sin(4x) +cos(4x)) +cos(x)(sin(4x) +cos(4x))
= (sin(x) +cos(x))·(sin(4x) +cos(4x))
Each of the sums in this product is of the same form, so each can be simplified using the identity ...
sin(x) +cos(x) = √2·sin(x +π/4)
Then the given equation can be rewritten as ...
cos(3x) +sin(5x) = 0
2·sin(x +π/4)·sin(4x +π/4) = 0
Of course sin(x) = 0 for x = n·π, so these factors are zero when ...
sin(x +π/4) = 0 ⇒ x = nπ -π/4
sin(4x +π/4) = 0 ⇒ x = (nπ -π/4)/4 = (4nπ -π)/16
The solutions are ...
x ∈ {(n-1)π/4, (4n-1)π/16} . . . . . for any integer n
