SOLUTION: Solve algebraically : 3 cos2x+cosx+1=0

Algebra ->  Trigonometry-basics -> SOLUTION: Solve algebraically : 3 cos2x+cosx+1=0       Log On


   

Question 941795: Solve algebraically :
3 cos2x+cosx+1=0

Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
+cos%282x%29+=2cos%5E2%28x%29-1
3cos2x+cosx+1=0
+3%282cos%5E2%28x%29-1%29%2Bcosx%2B1=0
++6cos%5E2%28x%29+-3+%2Bcosx%2B1=0
+6cos%5E2%28x%29%2Bcos%28x%29+-2=0
let assume cos(x) =t
the above equation becomes as follows
+6t%5E2%2Bt-2=0
6t%5E2%2B4t-3t-2=0
2t%283t%2B2%29-1%283t%2B2%29=0
+%283t%2B2%29%282t-1%29=0
possible solotutions
either (3t+2)=0 or (2t-1)=0
3t=-2 2t =1
t= -2/3 t= 1/2
Hence t= -2/3 or 1/2
but t= cos(x)
hence cos(x) = -2/3 or 1/2
x= 131.81 degrees or 60 degrees