Is it possible to have two solutions to this problem? Here is how I went about it. sin(x-6)=cos(3x-4) sin(30) = cos(60), therefore: x-6=30 and 3x-4=60 x=36 and 3x=64 x=64/3 Feb 5 | Michael J.
The equation has to work for each solution value. The solutions to the equation are the zeros of the graphed function; there are an infinite number of them. Since 6 and -4 do not have degree symbols, you have to work in radians. sin(30 radians) ≈ -0.988031624092862 ≠ cos(60 radians) ≈ -0.952412980415156 IF the equation were sin(x°-6°) = cos(3x°-4°): then if x°-6°=30°, x°=36°, and 3x°-4°=3*36°-4°=108°-4°=104°. sin(36°) ≈ 0.587785252292473 ≠ cos(104°) ≈ -0.241921895599668. Feb 5 | Steve S.
Comments
sin(x-6)=cos(3x-4)
sin(30) = cos(60), therefore:
x-6=30 and 3x-4=60
x=36 and 3x=64
x=64/3