precal
posted by john 2020 .
solve the equation on the interval [0,2pi)
cos(x+pi/4)cos(xpi/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(ab):
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