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


  • precal -

    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
    x = π + π/4 = 5π/4</b? or
    x = 2π - π/4 = 7π/4

  • check typing - precal -

    x = π + π/4 = 5π/4 or
    x = 2π - π/4 = 7π/4

  • precal -

    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

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