Question 622018: please help me solve for all:
2cos^2x-cosx=1
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 2cos^2x-cosx=1
2cos^2x-cosx=1
2cos^2x-cosx-1=0
2z^2-z-1=0 .... Let z = cosx
2z^2-2z+z-1=0
(2z^2-2z)+(z-1)=0
2z(z-1)+(z-1)=0
2z(z-1)+1(z-1)=0
(2z+1)(z-1)=0
(2z+1)(z-1)=0
2z+1=0 or z-1=0
z=-1/2 or z=1
cosx=-1/2 or cosx=1
x=arccos(-1/2) or x=arccos(1)
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x=arccos(-1/2)
x=2pi/3+2pi*n or x = 4pi/3 + 2pi*n
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x=arccos(1)
x=2pi*n
So the solutions are:
x=2pi/3+2pi*n, x = 4pi/3 + 2pi*n, or x=2pi*n
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