How to solve this trigonometric equation cos3x + sin5x = 0

I will be really glad, if you explain it step by step. Thanks!

  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

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .