cos(4x + π/5) - sin(2x) = 0
Since cos(x + π/2) = -sin(x), we can write
cos(4x + π/5) = cos(4x + π/2 - 3π/10) = -sin(4x - 3π/10)
Then we can absorb the negative signs into either sin expression, using the fact that sin(-x) = -sin(x):
-sin(4x - 3π/10) - sin(2x) = 0
sin(3π/10 - 4x) + sin(-2x) = 0
Recall that
• sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
• sin(a - b) = sin(a) cos(b) - cos(a) sin(b)
==> • 2 sin(a) cos(b) = sin(a + b) + sin(a - b)
We can further condense the left side in our equation if there exists a solution to
{ 3π/10 - 4x = a + b
{ -2x = a - b
We have
(a + b) + (a - b) = (3π/10 - 4x) + (-2x)
==> 2a = 3π/10 - 6x
==> a = 3π/20 - 3x
(a + b) - (a - b) = (3π/10 - 4x) - (-2x)
==> 2b = 3π/10 - 2x
==> b = 3π/20 - x
So in our equation, we have
sin(3π/10 - 4x) + sin(-2x) = 0
2 sin(3π/20 - 3x) cos(3π/20 - x) = 0
sin(3π/20 - 3x) cos(3π/20 - x) = 0
Solve for x :
sin(3π/20 - 3x) = 0 or cos(3π/20 - x) = 0
[3π/20 - 3x = arcsin(0) + nπ or 3π/20 - 3x = π - arcsin(0) + nπ]
or [3π/20 - x = arccos(0) + nπ or 3π/20 - x = -arccos(0) + nπ]
[3π/20 - 3x = nπ or 3π/20 - 3x = π + nπ]
or [3π/20 - x = π/2 + nπ or 3π/20 - x = -π/2 + nπ]
[3x = 3π/20 - nπ or 3x = -17π/20 - nπ]
or [x = -7π/20 - nπ or x = 13π/20 - nπ]
[x = π/20 - nπ/3 or x = -17π/60 - nπ/3]
or [x = -7π/20 - nπ or x = 13π/20 - nπ]
(where n is any integer)
