+1 vote
in JEE by kratos

The number of solution of cosx + cos2x + cos3x + cos4x = 0, x ∈ [0, 2π] is

(a) 4

(b) 5

(c) 6

(d) 7

1 Answer

+4 votes
by kratos
 
Best answer

The correct option (d) 7

Explanation:

∵ cosx + cos2x + cos3x + cos4x = 0

∴ cosx + cos3x + cos2x + cos4x = 0

∴ 2cos2x ∙ cosx + 2cos3x ∙ cosx = 0

∴ 2cosx(cos2x + cos3x) = 0

∴ 4cosx ∙ cos(5x/2) ∙ cos(x/2) = 0

∴ x = (π/2), (3π/2), (π/5), (3π/5), (7π/5), (9π/5) & π

∴ Total solution = 7.

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