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Homework Statement
3cos(x)+4sin(x)=5
Solve for X.
Homework Equations
acos(x)+bsin(x)=c
f(x)=Rsin(x+θ)
The Attempt at a Solution
Rsin(x+θ)=5
Rsin(x)cos(θ)+Rcos(x)sin(θ)=5
Rcos(θ)=4
Rsin(θ)= 3
R^2*cos^2(θ)+R^2*sin^2(θ)=R^2=3^2+4^2=25
R=+/-2
5sin(x+θ)=5
sin(x+θ)=1
x+θ=pi/2+/-2npi
x=pi/2-θ+/-2npi
Rsin(θ)/Rcos(θ)=tan(θ)=3/4
arctan(3/4)=0.64 rad
so X=pi/2-0.64+/-2npi
What confuses me is that I get a different result if I use negative R:
x=-pi/2-θ+/-2npi
Are -pi/2 and pi/2 equal?
Thank You.