Question

Cos(4x + pi/5) - sin 2x = 0

Answer 1

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)