solve the equation on the interval [0,2pi)

cos(x+pi/4)-cos(x-pi/4)=1

first expand it

cosxcosπ/4 - sinxsinπ/4 - (cosxcoxπ/4 + sinxsinπ/4) = 1
-2sinxsinπ/4 = 1
-2sinx(√2/2) = 1 , since we know sinπ/4 = 1/√2 or √2/2
sinx = -1/√2
the sine is negative in III and IV
so
x = π + π/4 = 5π/4</b? or
x = 2π - π/4 = 7π/4

x = π + π/4 = 5π/4 or

x = 2π - π/4 = 7π/4

Use the identity formulas for cos(a+b) and cos(a-b):

cosx*cos(pi/4) - sinx*sin(pi/4)-cosx*cos(pi/4) - sinx*sin(pi/4) = 1
-2 sinx*sin(pi/4)= 1
sinx (1/sqrt2)= -1/2
sinx = -(sqrt2)/2
x = 5 pi/4 or 7 pi/4

To solve the equation cos(x+π/4) - cos(x-π/4) = 1 on the interval [0, 2π), we can use the trigonometric identity for the difference of two cosines:

cos(A) - cos(B) = -2sin((A+B)/2)sin((A-B)/2)

Using this identity, we can rewrite the equation as follows:

-2sin((x+π/4 + x-π/4)/2)sin((x+π/4 - x+π/4)/2) = 1

Simplifying further:

-2sin(x/2)sin(π/4) = 1

sin(x/2) = -1 / (2√2)

Now, we need to find the values of x/2 that satisfy this equation. To do that, we can take the inverse sine (arcsin) of both sides to isolate x/2:

x/2 = arcsin(-1 / (2√2))

Since we're looking for the solutions on the interval [0, 2π), we can restrict the values of x/2 accordingly.

The arcsin function has a range of [-π/2, π/2], so the negative value on the right side of the equation corresponds to an angle in the 4th quadrant. To find the positive angle in the 1st quadrant, we can add 2π to the negative angle:

x1 = 2π + arcsin(-1 / (2√2))

Similarly, we can find the negative angle in the 3rd quadrant by subtracting the negative angle from 2π:

x2 = 2π - arcsin(-1 / (2√2))

Therefore, the solutions to the equation on the interval [0, 2π) are x1 and x2.