Question

No of solutions of equation Cosx+cos2x+cos3x =0 in x belongs to 0 to pi

Answer 1

cosx + cos2x + cos3x = 0 like $x\in[0,\pi]$

cosx + cos3x + cos2x = 0.

2cos(x + 3x)/2. cos(3x - x)/2 + cos2x = 0

[ from cosC + cosD = 2cos(C + D)/2.cos(C - D)/2]

2cos2x.cosx + cos2x = 0

cos2x(2cosx + 1) = 0

cos2x = 0, cosx = -1/2

when cos2x = 0 = cosπ/2
2x = 2nπ ± π/2 , where n is integers
x = nπ ± π/4
if n = 0, x = ± π/4 , but $x\in[0,\pi]$ so, x = π/4
if n = 1 , x = π ± π/4, x = 5π/4 , 3π/4 but x ≠ 5π/4
hence, x = π/4 and 3π/4 from cos2x = 0

when cosx = -1/2 = cos2π/3
x = 2nπ ± 2π/3
for $x\in[0,\pi]$ only x = 2π/3 is solution

hence, there are three solutions .e.g., x = π/4, 3π/4 and 2π/3

Answer 2

Answer:cosx + cos2x + cos3x = 0 like

cosx + cos3x + cos2x = 0.

2cos(x + 3x)/2. cos(3x - x)/2 + cos2x = 0

[ from cosC + cosD = 2cos(C + D)/2.cos(C - D)/2]

2cos2x.cosx + cos2x = 0

cos2x(2cosx + 1) = 0

cos2x = 0, cosx = -1/2

when cos2x = 0 = cosπ/2

2x = 2nπ ± π/2 , where n is integers

x = nπ ± π/4

if n = 0, x = ± π/4 , but so, x = π/4

if n = 1 , x = π ± π/4, x = 5π/4 , 3π/4 but x ≠ 5π/4

hence, x = π/4 and 3π/4 from cos2x = 0

when cosx = -1/2 = cos2π/3

x = 2nπ ± 2π/3

for only x = 2π/3 is solution

hence, there are three solutions .e.g., x = π/4, 3π/4 and 2π/3