Can you solve this trig equation

Hey guys,

Please try and find the general solution to:

cos3xsinx = cos4xsin2x

Thank you

⇒ cos(3x)sin(x) = cos(4x)sin(2x)
⇒ cos(3x)sin(x) - cos(4x)sin(2x) = 0

Use double angle formula: sin(2x) = 2 sin(x) cos(x)

⇒ cos(3x)sin(x) - cos(4x)(2sin(x)cos(x)) = 0
⇒ sin(x) [cos(3x) - 2 cos(4x)cos(x)] = 0

Use product-to-sum formula: cos(A)cos(B) = 1/2 (cos(A+B) + cos(A-B))

⇒ sin(x) [cos(3x) - 2 * 1/2 (cos(5x) + cos(3x))] = 0
⇒ sin(x) [cos(3x) - (cos(5x) + cos(3x))] = 0
⇒ -sin(x) cos(5x) = 0

sin(x) = 0
x = kπ

cos(5x) = 0
5x = π/2 + kπ
x = π/10 + kπ/5


SOLUTION: x = kπ, π/10 + kπ/5
for all integers k

[sin2x = 2sinxcosx] [double-angle formula]

Divide both sides by sinx
** By doing this we assume that sinx ≠ 0.
** If sinx=0, then the statement is true since sin2x=0 also, so 0=0.
** So x=arcsin(0) = aπ is a valid solution for all integers a.

cos3x = 2cosxcos4x

[cosxcos4x = (cos3x + cos5x)/2] [product-to-sum formula]

cos3x = cos3x + cos5x

cos5x = 0.

5x = arccos(0)
5x = ±π/2, ±3π/2, … (2a+1)π/2

x = (2a+1)π/10

So the solutions are:

x = (2a+1)π/10
OR
x = aπ

@ ICU: You sure you got the right question?

3, 15, 19, e, 24, 19, 9, 14, 24 = 3, 15, 19, d, 24, 19, 9, 14, b, 24