qqmore said:
Find the general solution of the equation cosx + cos2x + cos3x = 0.
Can this be done other then using the summation to product formula?
cosx + cos2x+ cos3x = cosx + 2cos^2(x) -1 + cos(2x+x)
= cosx +2cos^2(x) - 1 + cos2xcosx-sin2xsinx
= cosx +2cos^2(x) - 1 +(2cos^2(x)-1) . cosx - (2sinxcosx) . sinx
= cosx + 2cos^2(x) - 1 + 2cos^3(x) - cos x - 2sin^2(x) cosx
= 2cos^2(x) -1 + 2cos^3(x) - 2(1-cos^2(x)) . cosx
= 2cos^2(x) - 1 + 2cos^3(x) - 2cosx +2cos^3(x)
= 4cos^3(x) + 2cos^2(x) - 2cosx - 1
= (2cos^2(x)-1)(2cosx + 1)
Since cosx + cos2x + cos3x = 0, (2cos^2(x)-1)(2cosx + 1) = 0
Therefore, 2cos^2(x) = 1 or 2cosx = -1
cos^2(x) = 1/2 or cosx = -1/2
Then cosx = 1/(square root of 2), -1/(square root of 2) or -1/2
x = pi/4, 3pi/4 or 2pi/3
= 2npi +_ (i.e. plus & minus) pi/4, 2npi +_ 3pi/4 or 2npi +_ 2pi/3
You need to keep factorising to get the values of x. Once you get them, you can use the general solution formula for cos to find the answer.
General solution formula for cos = 2npi +_ x