The number of solutions of the equation `1+cosx+cos2x+sinx+sin2x+sin3x=0,` which satisfy the condition `pi/2<|3x-pi/2|lt=pi` is

Answer

2

Explanation

Given `pi/2 lt |3x - pi/2| lt pi`
`rArr pi/2 lt (3x - pi/2) le pi`
or `-pi le (3x - pi/2) lt (-pi)/2`
`rArr x in [(-pi)/6, 0) uu (pi/3, pi/2]`
Now, `1+cos x + cos 2x + sin x + sin 2x + sin 3x =0`
`rArr 2 cos^(2) x+ cos x + sin 2x +2 sin 2x cos x=0`
`rArr cos x(2 cos x +1) + sin 2x (2 cos x + 1) =0`
`rArr (cos x + sin 2x) (2 cos x+1)=0`
`rArr cos x(1+2 sin x) (2 cos x+1) =0`
`rArr cos x=0` or `sin x = (-1)/2` (as for given interval, `cos x gt 0`)
`rArr x=pi/2` or `x= (-pi)/6`
Hence, there are 2 solutions.