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1.

Let n be odd integer . If `sin theta = sum_(r=0)^(n) b_(r) sin^(r) theta`, for every value of `theta`, thenA. `b_(0) = 1, b_(1) =3`B. `b_(0) = 0, b_(1) `C. `b_(0) = - 1 , b_(1) =n `D. `b_(0) =0, b_(1) = n^(2) - 3n +3`

Answer» Correct Answer - B
Given, `sin n theta = underset(r=0)overset(n)sum b_(r) sin^(r) theta`
Now put, `theta = 0`, we get `0 =b_(0)`
`therefore sin ntheta =underset(r=1)overset(n) b_(r) sin^(r) theta`
`rArr (sin n theta)/(sin theta) = underset(r=1)overset(n) sum b_(r) (sin theta)^(r-1)`
Taking limit as ` theta to0`
`rArr underset( theta to 0)(lim) (sin ntheta)/(sin theta) = underset(theta to 0)(lim) underset(r=1)overset(n)sum b_(r) (sin theta)^(n-1)`
`rArr underset(theta to 0)(lim) (n theta.(sin ntheta)/(n theta))/(theta.(sintheta)/(theta)) = b_(1) + 0 + 0 + 0.....`
[`because` other values become zero for higher power of sin `theta`]
`rArr " "(n.1)/(1) =b_(1)`
`rArr " "b_(1) = n`
Discussion
2.

`sec^2 x=(4xy)/(x+y)^2` is true if and only ifA. `x = y ne 0 `B. `x = y,x ne 0`C. `x = y`D. `x ne 0, y ne 0`

Answer» Correct Answer - (a,b)
We know that, ` sec ^(2) theta ge 1 `
` rArr " " ( 4xy )/((x + y ) ^(2)) ge 1 `
` rArr " " 4xy ge (x + y ) ^(2)`
` rArr ( x + y) ^(2) - 4xy le k 0 `
` rArr " " (x - y )^(2) le 0 `
` rArr x - y = 0 `
` rArr " " x = y `
Therefore, ` x + y = 2x " " ` [add x both sides]
But ` x + y ne 0 ` since it lies in the denominator,
`rArr 2x ne 0 `
`rArr x ne 0 `
Hence, ` x = y , x ne 0 ` is the answer.
Therefore, (a) and (b) are the answers.
Discussion
3.

The number of all the possible triplets `(a_1,a_2,a_3)`such that `a_1+a_2cos(2x)+a_2sin^2(x)=0`for all `x`is0 (b)1 (c) 3(d) infinite

Answer» Correct Answer - d
Given, ` a_1 + a_2 cos 2x + a _3 sin ^(2) x = 0, AA x `
` rArr a _1 + a_2 cos 2x + a _3 ((1 - cos 2x)/(2x)) = 0, AA x `
` rArr (a_1 + (a_3)/(2)) + (a_2 - (a_3)/(2)) cos 2 x = 0, AA x `
` rArr a_1 + (a_3)/(2) = 0 and a_2 - (a_3)/(2) = 0 `
` rArr a _1 = - (k)/(2), a_2 = (k)/(2) , a_3 = k `, where `k in R`
Hence, the solutions, are ` (-(k)/(2), (k )/(2), k)`, where k is any real number.
Thus, the number of triplets is infinite.
Discussion
4.

The smallest positive value of x (in degrees) for which`tan(x+100^@)=tan(x+50^@).tanx.tan(x-50^@)` is

Answer» Correct Answer - `x = 30^(@)`
`tan(x+100^(@))=tan(x+50^(@))tanxtan(x-50^(@))`
`implies(tan(x+100^(@)))/(tanx)=tan(x+50^(@))tanxtan(x-50^(@)).`
`implies(sin(x+100^(@)))/(cos(x+100^(@))).(cosx)/(sinx)=(sin(x+50^(@))sin(x-50^(@)))/(cos(x+50^(@))cos(x-50^(@)))`
`implies(sin(2x+100^(@))+sin100^(@))/(sin(2x+100^(@))-sin100^(@))=(cos100^(@)-cos2x)/(cos100^(@)+cos2x)`
`implies[sin(2x+100^(@))+sin100^(@)][cos100^(@)+cos2x]`
`=[cos100^(@)-cos2x]xx[sin(2x+100^(@))-sin100^(@)]`
`impliessin(2x+100^(@)).cos100^(@)+sin(2x+100^(@)).cos2x+sin100^(@)cos100^(@)+sin100^(@)cos2x`
`=cos100^(@)sin(2x+100^(@))-cos100^(@)sin100^(@)-cos2xsin(2x+100^(@))+cos2xsin100^(@)`
`implies2sin(2x+100^(@))cos2x+2sin100^(@)cos100^(@)=0`
`impliessin(4x+100^(@))+sin100^(@)+sin200^(@)=0`
`=sin(4x+100^(@))+2sin150^(@)cos50^(@)=0`
`=sin(4x+100^(@))+2.(1)/(2)sin(90^(@)-50^(@))=0`
`impliessin(4x+100^(@))+sin40^(@)=0`
`impliessin(4x+100^(@))=sin(-40^(@))`
`implies4x+100^(@)=npi+(-1)^(n)(-40^(@))`
`implies4x=n(180^(@))+(-1)^(n)(-40^(@))-100^(@)`
`impliesx=(1)/(4)[n(180^(@))+(-1)^(n)(-40^(@))-100^(@)]`
The smallest positive value of x is obtained when n=1.
Therefore, `x=(1)/(4)(180^(@)+40^(@)-100^(@))`
`impliesx=(1)/(4)(120^(@))=30^(@)`
Discussion
5.

The positive integer value of `n >3`satisfying the equation`1/(sin(pi/n))=1/(sin((2pi)/n))+1/(sin((3pi)/n))i s`

Answer» Correct Answer - 7
Given, ` n gt3 in `Integer
` and (1)/(sin ((pi)/(n ))) = (1)/(sin ((2pi)/(n))) + (1)/(sin ((3pi)/(n)))`
` rArr (1)/(sin ""(pi)/(n)) - (1)/(sin ""(3pi)/(n)) = (1)/(sin ""(2pi)/(n)) `
`rArr ( sin ""( 3pi)/(n ) -sin ""( pi )/(n))/(sin ""(pi)/(n)* sin ""(3pi)/(n)) = (1)/(sin ""( 2pi )/(n))`
`rArr 2 cos (( 2pi )/( n )) * sin ""(pi)/(n) = ( sin ""(pi)/(n) * sin ""(3pi )/(n))/( sin ""( 2pi )/(n )) `
` rArr 2 sin ""(2pi)/(2)* cos ""( 2pi )/(n) = sin ""(3pi)/(n)`
`rArr " " sin ""(4pi)/(n) = sin ""( 3pi )/(n) `
` rArr " " ( 4pi)/(n ) = pi - ( 3pi )/(n)`
` rArr " " ( 7pi)/(n) = pi rArr n = 7`
Discussion
6.

The smallest positive root of the equation `tanx-x=0`lies in`(0,pi/2)`(b) `(pi/2,pi)``(pi,(3pi)/2)`(d) `((3pi)/2,2pi)`A. `(0,(pi)/(2))`B. `((pi)/(2),pi)`C. `(pi,(3pi)/(2))`D. `((3pi)/(2),2pi)`

Answer» Correct Answer - c
Let ` f (x) = tan x - x `
We know, for ` 0 lt x lt (pi)/(2)`
` rArr " " tanx gt x `
` therefore " " f (x) = tan x - x ` has no root in ` ( 0 , pi//2)`
For ` pi//2 lt x lt pi , tan x ` is negative.
` therefore " " f (x) = tan x - x lt 0 `
So, ` " " f(x) = 0 ` has no root in ` ((pi)/(2), pi )`
For ` " " ( 3pi )/(2) lt x lt 2pi, tan x ` is negative.
` therefore " " f (x) = tan x - x lt 0 `
So, ` f (x) = 0` has no root in ` (( 3pi)/(2), 2pi)`
We have, ` f(pi) = 0 -pi lt 0 `
and ` f((3pi)/(2)) = tan ""(3pi)/(2) - ( 3pi)/(2) gt 0 `
` therefore f (x) =0` has at least one root between `pi and ( 3pi )/(2)`
Discussion
7.

The set of all `x`in the interval `[0,pi]`for which `2sin^2x-3sinx+1geq0`is______

Answer» Correct Answer - `x in [0,(pi)/(6)]uu{(pi)/(2)} uu [(5pi)/(6),pi]`
Given, ` 2 sin ^(2) x - 3 sin x + 1 ge 0`
` rArr 2 sin ^(2) x - 2 sin x - sin x + 1 ge 0 ` ltBrgt ` rArr " " ( 2 sin x - 1 ) (sin x - 1 ) ge 0 `
` rArr 2 sin x - 1 le 0 or sin x ge 1 `
` rArr " " sin x le (1)/(2) or sin x = 1 `
` rArr " "x in [0, (pi)/(6)] uu {(pi)/(2)} uu [ (5pi)/(6), pi]`
Discussion
8.

Find the smallest positive number p for which the equation `cos (p sin x) = sin (p cos x)` has a solution `x in [0,2pi]`.

Answer» Correct Answer - Smallest positive value of `p =(pi)/(2sqrt(2))`
Given, ` cos (p sinx ) = sin(p cos x ), AA x in [0, 2pi]`
` rArr cos (p sin x ) = cos ( (pi)/(2) - p cos x) `
` rArr" " p sin x = 2 npi pm ((pi)/(2) - p cos x ) , ne in I `
` " "[because cos theta = cos alpha rArr theta = 2n pi pm alpha, n in I ] `
` rArr " " p sinx + p cos x = 2 npi + pi//2 `
or ` " " p sinx - p cos x = 2n pi - pi//2, n in I `
` rArr " " p (sin x + cos x ) = 2 n pi + pi//2`
or ` " " p (sinx - cos x ) = 2 n pi - pi //2, n in I `
` rArr p sqrt 2 (cos ""(pi)/(4) sinx + sin ""(pi)/(4) cos x) = 2 n pi + (pi)/(2)`
or ` psqrt2(cos ""(pi)/(4) sin x - sin ""(pi)/(4) cos x ) = 2n pi - (pi)/(4) , n in I `
`rArr " " psqrt 2[sin(x - pi//4)] = ((4n + 1)pi )/(2)`
` or " " psqrt2[sin(x - pi//4)] = ( 4n - 1 ) (pi)/(2), n in I`
Now, `" " - 1 le sin (x + pi//4) le 1 `
` rArr" " - psqrt2 le psqrt2 sin ( x pm pi//4) le p sqrt2`
`rArr " " -psqrt2 le ((4n + 1 )*pi)/(2) le p sqrt2, n in I`
or ` " "-psqrt2 le ((4n - 1 )pi)/(2) le psqrt2 , n in I `
Second inequality is always a subset of first, therefore we have to considor only first.
It is sufficient to consier `n ge 0 `, because for ` n gt 0`, the solution will be same for `n ge 0.`
If `n ge 0, " " - sqrt2p le ( 4n + 1) pi//2`
` rArr" " (4n + 1) pi//2 le sqrt2p `
For p to be least, n should be latest.
` rArr" " n = 0 `
` rArr " " sqrt2 p ge pi//2 rArr p ge (pi)/(2sqrt2)`
Therefore, least value of p = `(pi)/(2sqrt2)`
Discussion
9.

The solution set of the system of equations `x+y=(2pi)/3,cosx+cosy=3/2,`where `xa n dy`are real, is ________

Answer» Correct Answer - No solution
Given, ` x + y = ( 2pi)/(3)`
` and " " cos x + cos y = (3)/(2)`
` rArr cos x + cos (( 2pi)/(3) - x ) = (3)/(2)`
`rArr cos x + (-(1)/(2) cos x + (sqrt3)/(2) sinx ) = (3)/(2)`
` rArr " " (1)/(2) cos x + (sqrt3)/(2) sinx = (3)/(2)`
` rArr sin ((pi)/(6) + x ) = (3)/(2)` , which is never possible.
Hence, no solution exists.
Discussion
10.

Let S be the set of all `alpha in R` such that the equation, `cos 2x + alpha sin x = 2alpha -7` has a solution. The S, is equal toA. RB. `[1,4]`C. `[3,7]`D. `[2,6]`

Answer» Correct Answer - D
The given trigonometric equciton is `cos 2 x + alpha sin x = 2alpha -7`
`rArr 1-2 sin^(2) x + alpha x = 2alpha - 7 " "[therefore 2x = 1- 2 sin^(2) x]`
`rArr 2 sin^(2) - alpha sin x + 2alpha - 8 = 0 `
` rArr 2 (sin ^(2) x - 4) - alpha (sin x - 2 ) = 0`
` rArr 2 (sin x-2) (sin x + 2) -alpha(sin x - 2)=0`
`rArr (sin x-2) (2 sin x + 4-alpha) = 0 `
`therefore 2 sin x + 4 - alpha = 0" " [because sin x- 2 ne 0]`
`rArr sin x = (alpha-4)/(2)" ".....(i)`
Now, as we know `-1 le sin x le 1`
`therefore " " -1 le (alpha-4)/(2) le 1" "["from Eq.(i)"]`
`-2 le alpha - 4 le 2 rArr 2 le alpha le 6 rArr in [2,6]`
and range of ` cos^(2) (3x) = [0,1]`
So, the given equation holds if
`1 + sin ^(4) x = 1 cos^(2)(3x)`
`rArr sin^(4) x= 0 and cos^(2)3x =1`
Since, `x in [-(5pi)/(2),(5pi)/(2)]`
`therefore x = - 2pi, - pi,0, pi , 2pi`
Thus, there are five different value of x is possible .
Discussion
11.

The larger of `cos (log theta) and log (cos theta)` if `pi/2 lt theta lt pi/2`, is

Answer» Correct Answer - `cos (log theta)`
Since, ` cos theta le 1 rArr log (cos theta ) lt 0`
and `" " cos ( log theta ) gt 0 `
` therefore cos ( log theta ) gt log (cos theta )`
Discussion
12.

If `alpha and beta` are non-zero real number such that `2(cos beta-cos alpha)+cos alpha cos beta=1.` Then which of the following is treu?A. `sqrt(3) tan ((alpha)/(2))-tan((beta)/(2))=0`B. `tan((alpha)/(2)) - sqrt(3) tan((beta)/(2)) = 0`C. `tan((alpha)/(2)) + sqrt(3)((beta)/(2))=0`D. `sqrt(3) tan((alpha)/(2))+ tan((beta)/(2))=0`

Answer» Correct Answer - B::C
We have, `2(cos beta - cos alpha)+ cos alpha cos beta = 1`
or `4(cos beta - cos alpha) + 2 cos alpha cos beta = 2`
`rArr 1-cos alpha + cos beta - cos alpha cos beta = 2`
`= 3+3 cos alpha - 3 cos beta - 3cos alpha cos beta`
`rArr (1-cos alpha) (1+ cos beta)=3(1+cos alpha) (1-cos beta)`
`rArr ((1-cos alpha))/((1+cos alpha)) = (3(1-cos beta))(1+cos beta))`
`rArr tan^(2) (alpha)/(2) = 3 tan^(2)(beta)/(2)`
`therefore tan.(alpha)/(2) pm sqrt(3) tan.(beta)/(2) =0`
Discussion
13.

If `sin^4 alpha + 4 cos^4 beta + 2 = 4sqrt2 sin alpha cos beta ; alpha, beta in [0,pi],` then `cos(alpha+beta) - cos(alpha - beta)` is equal to :A. `-1`B. `sqrt(2)`C. `-sqrt(3)`D. 0

Answer» Correct Answer - C
By applying `AM ge GM` inequality, on the numbers
`sin^(4) alpha, 4 cos^(2) beta`, 1 and 1, we get
`(sin^(4) alpha + 4 cos^(2) beta + 2)/(4)le (( sin^(4) alpha )(4cos^(4) beta).1.)^(1//4)`
`rArr sin^(4) alpha + 4 cos^(4) Beta + 2 ge 4sqrt(2)sin alpha cos beta`
But, it is given that
`sin^(4) alpha + 4 cos^(4) beta + 2 = 4sqrt(2)sin alpha cos beta`
So, `sin^(4) alpha= 4 cos ^(4) beta =1`
`[because In AM ge GM`, equality holds when all given positive quantites are equal.]
`rArr sin alpha = 1` and `sin beta = (1)/(sqrt(2)) " "......(i)`
Now, `cos (alpha + beta) -cos (alpha + beta)= - 2sin alpha sin beta " "[because alpha, beta in [0, pi]]`
`[because cos C - cos D = 2sin .(C + D)/(2)sin .(D-C)/(2)]`
`= -2 xx 1 xx (1)/(sqrt(2)) " "["From Eq. (i)"]`
`= - sqrt(2)`
Discussion
14.

Let `alpha` and `beta` be the roots of the quadratic equation `x sin^(2) theta -x (sin theta cos theta + 1) + cos theta =0 (0le thetale 45^(@)) and alpha le beta` . Then`sum_(n=0)^(oo) (alpha^(n)+((-1)^(n))/(beta^(n)))` is to equal toA. `(1)/(1-cos theta) - (1)/(1+sin theta)`B. `(1)/(1-cos theta)+(1)/(1+sin theta)`C. `(1)/(1+cos theta) - (1)/(1-sin theta)`D. `(1)/(1+cos theta)+(1)/(1+sin theta)`

Answer» Correct Answer - B
Given
`x^(2) sin theta -x sin theta cos theta -x + cos theta = 0`
Where `0 lt theta lt 45^(@)`
`sin theta (x- cos theta) - 1( x - costheta) = 0`
`rArr (x- cos theta)(x sin theta-1) = 0`
`rArr x = cos theta, x cosec theta`
`x = cos theta, x = coses theta`
`rArr alpha cos theta and beta = cosec theta`
`(because "For" 0 lt theta lt 40^(@), (1)/(sqrt(2)) lt cos theta lt sqrt(2) lt "cosec " theta lt oo rArr cos theta lt cosec theta)`
Now, consider, `underset(n = 0)overset(oo)sum(alpha^(n)+ ((-1)^(n))/(beta^(n))) = underset(n = 0 )overset(oo)sum + underset(n=0) overset(oo)sum ((-1)^(n))/(beta^(n))`
`= (1 + alpha + alpha^(2) + alpha^(3)+........oo)`
`+ (1-(1)/(beta) +(1)/(beta^(2)) - (1)/(beta^(3)) +.....oo)`
`= (1)/(1-alpha) + (1)/(1-((-1)/(beta)))= (1)/(1-alpha) +(1)/(1+(1)/(beta))`
`= (1)/(1-cos theta) +(1)/(1+ sin theta) " "{because (1)/(beta)sin^(2) = 1-cos^(2)x}`
Discussion
15.

If `0 le x le (pi)/(2)`, then the number of value of x for which `sin x - sin 2x + sin 3x = 0`, isA. 2B. 3C. 1D. 4

Answer» Correct Answer - A
We have , `sin x - sin 2x + sin 3x = 0`
`rArr 2 sin ((x + 3x)/(2)) cos((x-3x)/(2)) - sin 2x = 0`
`[because sin C + sin D = 2 sin((C+D)/(2)) cos ((C-D)/(2))]`
`rArr 2 sin 2x cos x - sin 2x = 0" "[because cos (-theta) = cos theat]`
`rArr sin 2x (2cos x-1)=0`
`rArr sin 2x=0 or 2 cos x - 1 =0`
`2x = 0, pi,......or cos x =(1)/(2)`
`rArr x = 0, (pi)/(2) ....or x =(pi)/(3)`
In the interval `[0,(pi)/(2))` only tow values satisfly , namely `x = 0` and ` = (pi)/(3)`.
Discussion
16.

The sum of all vaues of `theta in (0,(pi)/(2))` satisfying `sin^(2) x - sin 2x + sin 3x = 0`,isA. `(3pi)/(8)`B. `(5pi)/(4)`C. `(pi)/(2)`D. `pi`

Answer» Correct Answer - C
Given, `sin^(2) 2 theta + cos ^(4) 2 theta = (3)/(4)`
`rArr (1-cos^(2) 2 theta) + (cos^(4) 2 theta = (3)/(4) " "(because sin^(2) x = 1 - cos^(2) x)`
`rArr 4 cos^(4) 2 theta - 4 cos^(2) theta +1= 0`
`rArr ( 2 cos ^(2) 2theta - 1)^(2) = 0`
`rArr 2 cos 2 theta -1 = 0`
`rArr cos^(2) 2 theta = (1)/(2)`
`rArr cos 2 theta = pm (1)/(sqrt(2))`
If `theta in (0,(pi)/(2))`, then `2 theta in (0,pi)`
`thereforecos 2 theta = (pi)/(sqrt(2))`
`rArr 2 theta = (pi)/(4),(3pi)/(2), [ because cos((3pi)/(4))= cos (pi-(pi)/(4)) = - cos .(pi)/(4) = -(1)/(sqrt(2))]`
`rArr theta = (pi)/(8),(3pi)/(8)`
Sum of value of `theta= (pi)/(8) + (3pi)/(8) = (pi)/(2)`
Discussion
17.

Q. Let `P={theta:sin theta-cos theta=sqrt2 cos theta}` and `Q={theta:sin theta+cos theta=sqrt2 sin theta}` be two sets. thenA. `P sub Q and Q-P ne phi`B. `Q cancel(sub) P`C. `P cancel(sub)Q`D. `P = Q`

Answer» Correct Answer - D
`P = {theta: sin theta -costheta = sqrt(2) cos theta}`
`rArr cos theta (sqrt(2)+1) = sin theta`
` rArr tan theta = sqrt(2)+1`
`rArr Q = { theta : sin theta + cos theta} = sqrt(2) sin theta`
`rArr sin theta (sqrt(2) -1) = cos theta`
`rArr tan theta =(1)/(sqrt(2-1)) xx (sqrt(2)+1)/(sqrt(2)+1) = (sqrt(2)+1)`
`therefore P =Q`
Discussion
18.

The number of distinct solutions of the equation `5/4cos^(2)2x + cos^4 x + sin^4 x+cos^6x+sin^6 x =2` in the interval `[0,2pi] ` is

Answer» Correct Answer - -8
Here, `(5)/(4) cos^(2)2x + (cos^(4)x +sin^(4)x) + (cos^(6)x + sin^(6)x) =2`
`rArr (5)/(4) cot2x +[(cos^(2)x + sin^(2) x)^(2) -2sin^(2) x cos^(2)x]+(cos ^(2)x + sin^(2)x) [(cos^(2)x + sin^(2)x)^(2)-3sin^(2) cos^(2) x]=2`
`rArr (5)/(4)cos^(2)x + (1-2sin^(2)x cos^(2)x) + (1-3cos^(2) xsin^(2)x)=2`
`rArr (5)/(4)cos^(2)2x-5sin^(2)x cos^(2)x = 0`
`rArr (5)/(4)cos^(2)2x-(5)/(4) sin^(2)2x=0`
`rArr (5)/(2)cos^(2)2x = (5)/(4)rArr cos^(2) 2x= (1)/(2)`
`rArr 2cos^(2)2x =1`
`rArr 1+ cos 4x =1`
`rArr cos 4x =0 as 0 le x le 2pi`
`therefore 4x={(pi)/(2),(3pi)/(8),(5pi)/(2),(7pi)/(2),(9pi)/(2),(11pi)/(2),(13pi)/(2),(15pi)/(2)}`
as ` 0 le 4x le 8 pi`
`rArr x = {(pi)/(8),(3pi)/(8),(5pi)/(8),(9pi)/(8),(11pi)/(8),(13pi)/(8),(15pi)/(8)}`
Hence ,the total number of solutions is 8.
Discussion
19.

Let `S={xepsilon(-pi,pi):x!=0,+pi/2}`The sum of all distinct solutions of the equation `sqrt3secx+cosecx+2(tan x-cot x)=0` in the set S is equal toA. `-(7pi)/(9)`B. `-(2pi)/(9)`C. 0D. `(5pi)/(9)`

Answer» Correct Answer - C
Give `(sqrt(3)) sec x + coses x + 2 (tan x - cot c) = 0`
`(-pi lt x lt pi) - {0,pm pi//2}`
`rArr sqrt(3) x cos x + 2 (sin^(2) x - cos^(2)x) = 0`
`rArr sqrt(3) sin x + cos x - 2 cos 2 x =0`
Multiplying and dividing by `sqrt(a^(2) + b^(2)) i.e., sqrt(3+1) = 2`.
We get
`2((sqrt(3))/(2) sin x + (1)/(2)cosx)- 2 cps 2x = 0`
`rArr (cos x cos .(pi)(3) + sin x.sin.(pi)/(2))-cos 2 x cos = 0 cos (x-(pi)/(3)) = cos2x`
`therefore " " 2x = 2npi+x -(pi)/(3)" " [underset(rArr theta = 2npipmalpha)"Since",cos theta =cos alpha] `
`rArr 2x = 2npi + x- (pi)/(3)`
`or " " 3x = 2xpi-x + (pi)/(3)`
`rArr" "x = 2npi - (pi)/(3)`
`or " " 3x = 2npi + (pi)/(3)`
`rArr x = 2npi - (pi)/(3)`
`or " " x= (2npi)/(3) + (pi)/(9)`
`therefore" " x= (-pi)/(3)`
` or" "x=(pi)/(9),(-5pi)/(9) ,(7pi)/(9)`
Now, sum of all distinct solutions `= (-pi)/(3) + (pi)/(9)-(5pi)/(9) + (7pi)/(9) = 0`
Discussion