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3 years ago

How to solve this trigonometric equation cos3x + sin5x = 0

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

5 0

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

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