sol.of eq:cosx+cos2x+cos3x+1/2=0is?

Question

1)2npi/7
2)npi/7
3)2npi+pi/7
4)none of these

Answer 1

Partial solution:
:Let's reduce it to a polynomial equation:
Trig identities: cos 2x = 2cos^2 x -1
cos 3x = cos x cos 2x -sin x sin 2x
= cos x(2 cos^2x -1) -sin x (2 sin x cos x)
which reduces to
4 cos ^3 x - 3 cos x
So we get
4 cos ^3 x + 2 cos^2 x -2cos x - 1/2 = 0
Now let u = cos x and clear fractions:
8 u^3 + 4u^2 -4u -1 = 0.
But this equation has no rational roots!
So, have I made a gross error somewhere or
does the problem really reduce to this cubic eqn? What
are the roots of this cubic eqn?

Answer 2

cosx+cos3x+cos2x+1/2=0
2cos2xcosx+cos2x+1/2=0
cos2x(2cosx+1)+1/2=0
(2cox^2x-1)(2cosx-1)+1/2=0
4cos^3x+2cos^2x-2cosx-1/2=0
8cos^3x+4cos^2x-4cosx-1=0
put cosx=t
8t^3+4t^2-4t-1=0
solve for 't' using your calculator and
calculate cosx

Answer 3

(4)

Answer 4

sorry
ans is 4)
bye