If cosx+2cos(x+240°)=1
Determine the general equations
Cos(x) + cos (x+240o) =1
⟹ \implies⟹
cos(x)+cos(x+24π18)=1cos(x)+cos(x+{24\pi \over18}) = 1cos(x)+cos(x+1824π)=1
cos(x)+cos(x+24π18)=cos(0)cos(x)+cos(x+{24\pi \over18}) = cos(0)cos(x)+cos(x+1824π)=cos(0)
We know that cos(a+b)=cos(a)cos(b)−sin(a)sin(b)\boxed{cos(a+b) = cos(a)cos(b)-sin(a)sin(b)}cos(a+b)=cos(a)cos(b)−sin(a)sin(b)
cos(x)+cos(x)cos(24π/18)−sin(x)sin(24π/18)=cos(0)\boxed{cos(x)+cos(x)cos(24\pi/18)-sin(x)sin(24\pi/18)=cos(0)}cos(x)+cos(x)cos(24π/18)−sin(x)sin(24π/18)=cos(0)
On solving this
We got
x=13(6πn+π)\boxed{x = \frac{1}{3}(6\pi n +\pi)}x=31(6πn+π)
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