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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
Solve `sin 2 theta+cos theta=0`. |
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Answer» We have `sin 2 theta+cos theta=0` or `cos theta=- sin 2 theta = cos (pi/2+2 theta)` `rArr theta=2 n pi pm (pi/2+2 theta) n in Z` Taking positive sign, we have `theta=2 npi + pi/2 +2 theta` `theta=2 n pi - pi/2, n in Z` Taking negative sign, we have `theta=2 n pi -(pi/2+2 theta)` `rArr theta=(2 n pi)/3-pi/6, n in Z` |
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| 202. |
Solve sec `4 theta- sec 2 theta=2`. |
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Answer» `1/(cos 4 theta)-1/(cos 2 theta)=2` or `cos 2 theta-cos 4 theta =2 cos 2 theta cos 4 theta = cos 2 theta + cos 6 theta` or `cos 6 theta + cos 4 theta=0` or `2 cos 5 theta cos theta =0` `rArr cos 5 theta=0" "or cos theta =0` `rArr 5 theta=(2n+1) pi/2 or theta=(2n+1) pi/2, n in Z` `rArr theta=(n+1/2) pi/5 or theta=(n+1/2) pi, n in Z` |
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| 203. |
If `sintheta=1/2a n dcostheta=-(sqrt(3))/2,`then the general value of `theta`is `(n in Z)dot`A. `2n pi+(5pi)/6`B. `2n pi+pi/6`C. `2n pi+(7pi)/6`D. `2npi+pi/4` |
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Answer» Correct Answer - A `sin theta=1//2` and `cos theta=-sqrt(3)//2` `rArr theta` lies in the second quadrant. `rArr sin theta=sin 5pi//6`, `:. Theta=2npi+(5pi//6), n in Z` |
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| 204. |
Sum of roots of the equation `x^2-2x^2sin^2(pix)/2+1=0`is |
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Answer» Correct Answer - A `x^(4)-2x^(2) "sin"^(2) pi/2 x+1=0` `rArr x^(4)+1=2x^(2) "sin"^(2) (pi x)/2` `rArr x^(4)+1=2x^(2) (1-"cos"^(2) (pi x)/2)` `rArr (x^(2)-1)^(2)+2x^(2) "cos"^(2) (pi x)/2=0` `rArr (x^(2)-1)=0` and `2x^(2) "cos"^(2) (pi x)/2=0` `rArr x= pm 1` |
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| 205. |
For `n in Z ,`the general solution of `(sqrt(3)-1)sintheta+(sqrt(3)+1)costheta=2i s(n in Z)``theta=2npi+-pi/4+pi/(12)``theta=npi+(-1)^npi/4+pi/(12)``theta=2npi+-pi/4``theta=npi+(-1)^npi/4-pi/(12)`A. `theta=2npi pm pi/4+pi/12`B. `theta=n pi +(-1)^(n) pi/4+pi/12`C. `theta=2npi pm pi/4`D. `theta=npi +(-1)^(n) pi/4- pi/12` |
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Answer» Correct Answer - A `(sqrt(3)-1) sin theta+(sqrt(3)+1) cos theta=2` or `((sqrt(3)-1))/(2sqrt(2)) sin theta+ ((sqrt(3)+1)/(2sqrt(2))) cos theta=1/sqrt(2)` or `"sin" pi/12 sin theta+"cos"pi/12 cos theta= "cos"pi/4` or `cos(theta-pi/12)="cos" pi/4` `rArr theta-pi/12=2npi pm pi/4, n in Z` or `theta=2npi pm pi/4+ pi/12, n in Z` |
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| 206. |
Solve `2 sin^(3) x=cos x`. |
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Answer» Correct Answer - `x= npi +pi/4, n in Z` Since `sin x=0` does not satisfy the equation, dividing the equation by `sin^(3)x` we get `cot x osec^(2) x=2` or `cot^(3)x+cot x-2=0` or `(cot x-1) (cot^(2) x+cot x+2)=0` or `cot x=1` `rArr x=npi+pi//4, n in Z` |
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| 207. |
Find the number of integral value of `n`so that `sinx(sinx+cosx)=n`has at least one solution. |
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Answer» `sin x(sin x+cos x)=n` or `sin^(2) x+ sin x cos x =n` or `(1- cos 2x)/2+(sin 2 x)/2 =n` or `sin 2x-cos 2x=2n -1` `rArr -sqrt(2) le 2n -1 le sqrt(2)` or `(1-sqrt(2))/2 le n le (1+sqrt(2))/2` or `n=0, 1` |
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| 208. |
The most general value for which `tantheta = -1 " and " costheta = 1/(sqrt(2)) " is " (n in Z)`A. `n pi+(7 pi)/4`B. `n pi+(-1)^(n) (7pi)/4`C. `2npi+(7pi)/4`D. none of these |
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Answer» Correct Answer - C Since `tan theta lt 0` and `cos theta gt 0, theta` lies in the fourth quadrant. Then `theta=7 pi//4`. Hence, the general value of `theta` is `2n pi+7 pi//4, n in Z`. |
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| 209. |
Solve `sqrt(3)` sec `2 theta=2`. |
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Answer» We have `sqrt(3) sec 2 theta=2` or `cos 2 theta=sqrt(3)/2=cos pi/6` `rArr 2theta=2npi pm pi/6, n in Z` or `theta=npi pm pi/12, n in Z` |
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| 210. |
Number of integral value(s) of `m`for which the equation `sinx-sqrt(3)cosx=(4m-6)/(4-m)`has solutions, `x in [0,2pi]`, is ________ |
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Answer» Correct Answer - 4 Since `-2 le sin x-sqrt(3) cos x le 2`, we have `-2 le (4m-6)/(4-m) le 2` or `-1 le (2m-3)/(4-m) le 1` If `(2m-3)/(4-m) le 1`, we have `((2m-3)-(4-m))/(4-m) le 0` or `(3m-7)/(m-4) ge 0` ...(i) Also, `-1 le (2m-3)/(4-m)` `rArr (m+1)/(m-4) le 0` From Eqs. (i) and (ii), we get `m in [-1, 7/3]` Therefore, the possible integers are `-1, 0, 1, 2`. |
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| 211. |
Solve that following equations :`"tantheta"+"tan"(theta+pi/3)+"tan"(theta+(2pi)/3)=3` |
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Answer» Here, we will use, `tan(A+B) = (tanA+tanB)/(1-tanAtanB)` It is given that , `tantheta+tan(theta+pi/3)+tan(theta+(2pi)/3) = 3` `=>tantheta+(tantheta+tan(pi/3))/(1-tanthetatan((pi)/3)) +(tantheta+tan((2pi)/3))/(1-tanthetatan((2pi)/3)) = 3` `=>(tantheta+sqrt3)/(1-sqrt3tantheta) +(tantheta-sqrt3)/(1+sqrt3tantheta) = 3-tantheta` `=>((tantheta+sqrt3)(1+sqrt3tantheta) +(tantheta-sqrt3)(1-sqrt3tantheta))/(1-3tan^2theta) = 3-tantheta` `=>(tantheta+sqrt3tan^2theta+sqrt3+3tantheta+tantheta-sqrt3tan^2theta-sqrt3+3tantheta)/(1-3tan^2theta) = 3-tantheta` `=>(8tantheta)/(1-3tan^2theta) = 3-tantheta` `=>8tantheta = (3-tantheta)(1-3tan^2theta)` `=>8tantheta = 3-9tan^2theta-tantheta+3tan^2theta` `=>9tantheta-3tan^3theta = 3-9tan^2theta` `=>3tantheta - tan^3theta =1-3tan^2theta` `=>(3tantheta - tan^3theta)/(1-3tan^2theta) = 1` `=>tan3theta = tan(pi/4)` `=>3theta = npi+pi/4` `=>theta = (npi)/3+pi/12` |
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| 212. |
Solve: `16^sin^(2x)16^cos^(2x)=10 ,0lt=x |
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Answer» `16^(sin^(2)x)+16^(1-sin^(2) x)=10` If `16^(sin^(2)x)=t`, then `t+16/t=10` Then Eq. (i) becomes `t^(2)-10t+16=0` or `t=2, 8` `rArr 16^(sin^(2) x)=16^(1//4)` or `16^(3//4)` `rArr sin x= pm 1/2, pm sqrt(3)/2` Now `sin x=1/2`, then `x=pi/6, (5pi)/6` `sin x=-1/2`, then `x=(7pi)/6` or `(11pi)/6` `sin x= sqrt(3)/2`, then `x=pi/3, (2pi)/3` `sin x= - sqrt(3)/2`, then `x=(4pi)/3, (5 pi)/3` Hence, there will be eight solutions in all. |
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| 213. |
The number of solutions of the equation `cos6x+tan^2x+cos6xtan^2x=1`in the interval `[0,2pi]`is4 (b) 5(c) 6 (d)7A. 4B. 5C. 6D. 7 |
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Answer» Correct Answer - D `cos 6x=(1-tan^(2) x)/(1+tan^(2) x) = cos 2x` `rArr 2 sin 2x. Sin 4x =0` `rArr sin x. cos x. sin 4x=0` `rArr x=n pi, 4pi = m pi, m, n in Z` `rArr x=n pi, x=(mpi)/4, m, n in Z` `:. x=0, pi/4, (3pi)/4, pi, (5pi)/4, (7pi)/4, 2pi` |
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| 214. |
If `=lt=xlt=2pi,`then the number of solutions of `3(sinx+cosx)-(sin^3x+cos^3x)=8`is0 (b) 1(c) 2 (d)4 |
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Answer» Correct Answer - A The given equation is `3 (sin x + cos x)-2 (sin x+ cos x) (1- sin x cos x)=8` or `(sin x+cos x) [3-2+2 sin x cos x]=8` or `(sin x+cos x)[sin^(2) x+cos^(2) x+2 sin x cos x]=8` or `(sin x+ cos x)^(3)=8` or `sin x + cos x =2` or `sin x, cos x=1`, which is not possible. Hence, the given equation has no solution. |
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| 215. |
Solve `sin x tan x -sin x+ tan x-1=0` for `x in [0, 2pi]`. |
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Answer» Correct Answer - `x= pi/4, (5 pi)/4` `sin x tan x-sin x+tan x-1=0` `rArr tan x(sin x+1)-(sin x +1)=0` `rArr (sin x+1) (tan x-1) =0` `rArr sin x=-1` or `tan x=1` For `sin x=-1, tan x` is not defined (as cos x=0). `:. tan x=1` `:. x=pi/4, (5pi)/4` |
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| 216. |
Solve `cot(x//2)-cosec (x//2)=cot x`. |
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Answer» Correct Answer - `x=4 n pi pm (2pi)/3, n in Z` `(cos (x//2))/(sin (x//2)) - (cos x)/(2 sin (x//2) cos (x//2)) = (1)/(sin (x//2))` `rArr 2 cos^(2) (x//2) - cos(x) = 2 cos(x//2)` `rArr 1 + cos x - cos x = 2 cos (x//2)` `rArr cos (x//2) = 1//2 = cos (pi//3)` `rArr x//2 = 2n pi -+pi//3, n in Z` `rArr x = 4n pi -+ 2 pi//3, n in Z` |
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| 217. |
The number of solution of `sec^(2) theta + cosec^(2) theta+2 cosec^(2) theta=8, 0 le theta le pi//2` isA. 4B. 3C. 0D. 2 |
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Answer» Correct Answer - D We have `1/(sin^(2) theta cos^(2) theta)+2/(sin^(2) theta)=8, sin theta ne 0, cos theta ne 0` or `1+2 cos^(2) theta =8 sin^(2) theta cos^(2) theta` or `1+2 cos^(2) theta =8 cos^(2) theta (1- cos^(2) theta)` or `8 cos^(4) theta-6 cos^(2) theta +1=0` or `(4 cos^(2) theta -1) (2 cos^(2) theta -1)=0` or `cos^(@) theta = 1//4= cos^(2) (pi//3)` or `cos^(2) theta=1//2=cos^(2) (pi//4)` or `theta=n pi pm (pi//3) or theta =npi pm (pi//4), n in Z` Hence, for `0 le theta le pi/2, theta = pi//3, theta =pi//4` |
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| 218. |
If `2 sin^(2) ((pi//2) cos^(2) x)=1-cos (pi sin 2x), x ne (2n + 1) pi//2, n in I`, then `cos 2x` is equal toA. `1//5`B. `3//5`C. `4//5`D. `1` |
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Answer» Correct Answer - B The given equation is equivalent to `2 sin^(2) ((pi//2) cos^(2) x)=2 sin^(2) ((pi//2) sin 2x)` or `cos^(2) x= sin 2x` or `cos x (cos x -2 sin x)=0` `rArr 1-2 tan x=0` as `cos x ne0, x ne (2n+1) pi/2` or `tan x=1/2` `rArr cos 2x=(1-tan^(2) x)/(1+tan^(2) x)=3/5` |
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| 219. |
Solve `tan 3 theta=-1`. |
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Answer» `tan 3 theta=-1=tan ((-pi)/4)` `rArr 3theta=npi+((-pi)/4), n in Z` or `theta=(npi)/3-pi/12, n in Z` |
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| 220. |
Solve `cosec^(2)theta-cot^(2) theta=cos theta`. |
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Answer» Correct Answer - No solution `cosec^(2) theta-cot^(2) theta= cos theta` `rArr cos theta=1` `rArr sin theta=0`, which is not possible as `cosec theta` and `cot theta` are not defined for this value of `sin theta`. |
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| 221. |
If `tan ((p pi)/4)=cot ((qpi)/4)`, then prove that `p+q=2(2n+1), n in Z`. |
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Answer» `tan((p pi)/4)=cot ((qpi)/4)=tan (pi/2-(qpi)/4)` `rArr (p pi)/4= n pi+pi/2-(q pi)/4` or `p/4=n+1/2-q/4` or `(p+q)/4=(2n+1)/2` or `p+q=2(2n+1), n in Z` |
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| 222. |
Solve `2 tan theta-cot theta=-1`. |
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Answer» `2 tan theta- cot theta=-1` or `2 tan theta- 1/(tan theta)=-1` or `2 tan^(2) theta+tan theta-1=0` or `(tan theta+1) (2 tan theta-1)=0` or `tan theta=-1 or tan theta=1/2` or `tan theta=tan ((-pi)/4) or tan theta=tan ("tan"^(-1) 1/2)` `rArr theta=n pi +((-pi)/4)` or `theta=mpi+alpha`, where `m, n in Z` and `tan alpha =1/2`. |
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| 223. |
Solve `tan 5 theta= cot 2 theta`. |
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Answer» `tan 5 theta=cot 2 theta=tan (pi/2-2 theta)` `rArr 5 theta=n pi +pi/2-2 theta` `rArr 7 theta=n pi +pi/2` `rArr theta=(n pi)/7+pi/14, ((2n+1)pi)/14`, where `n in Z`, but `n ne 3, 10, 17, ...` where `tan 5 theta` is not defined. |
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| 224. |
Solve `tan^(2) theta+cot^(2) theta=2`. |
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Answer» Correct Answer - `theta= n pi pm pi/4, n in Z` `(tan^(2) theta-1)^(2)=0` or `tan^(2) theta=1` `rArr theta = npi pm pi//4, n in Z` |
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| 225. |
Find the number of solution of `sin^2xcos^2x=1+cos^2xsin^4x`in the interval `[0,2pi]dot` |
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Answer» `sin^(2)x cos^(2)x=1+cos^(2)x sin^(4)x` `:. Sin^(2)x cos^(2)x(1-sin^(2) x)=1` `:. Sin^(2) x cos^(4)x=1` `:. Sin^(2)x=cos^(4)x=1`, which is not possible. |
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| 226. |
If `tan^(2) {pi(x+y)}+cot^(2) {pi (x+y)}=1+sqrt((2x)/(1+x^(2)))` where `x, y in R`, then find the least possible value of y. |
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Answer» Correct Answer - `1//4` `LHS ge 2, RHS le 2` Equality is possible The for `RHS, x=1` For `LHS, tan^(2) pi (1+y)=1` `:. tan^(2) pi y=1` Least value of y ia `1/4`. |
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| 227. |
The total number of ordered pairs `(x , y)`satisfying `|x|+|y|=2,sin((pix^2)/3)=1,`is equal to2 (b) 3(c) 4 (d)6A. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - C `|x|+|y|=2, sin ((pi x^(2))/3)=1` `rArr |x|, |y| in [0, 2], (pi x^(2))/3=(4n+1) pi/2, n in Z` `rArr x^(2)=(3(4n+1))/2=3/2`, as `|x| le 2` `rArr |a|=sqrt(3/2), |y|=4-sqrt(3/2)` Thus, there are four ordered pairs. |
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| 228. |
If `x , y in [0,2pi]`, then find the total number of ordered pairs `(x , y)`satisfying the equation `sinxcosy=1` |
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Answer» `sin x cos y=1` `rArr sin x=1, cos y=1 or sin x=-1, cos y =-1` If `sin x=1. cos y=1`, hence, `x=pi//2, y=0, 2pi` If `sin x=-1, cos y =-1`, hence, `x=3pi//2, y=pi` Thus, the possible ordered ordered pairs are `(pi/2, 0), (pi/2, 2pi)` and `((3pi)/2, pi)`. |
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| 229. |
`(sin^(3) theta-cos^(3) theta)/(sin theta - cos theta)- (cos theta)/sqrt(1+ cot^(2) theta)-2 tan theta cot theta=-1` ifA. `theta in (0, pi/2)`B. `theta in (pi/2, pi)`C. `theta in (pi, (3pi)/2)`D. `theta in ((3pi)/2, 2pi)` |
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Answer» Correct Answer - B `(sin^(2) theta+cos^(2) theta+ sin theta cos theta)- (cos theta)/(|cosec theta|)-2=-1` or `1+sin theta cos theta -|sin theta| cos theta-1=0` or `cos theta (sin theta -|sin theta|)=0` `rArr theta in (0, pi/2) or theta in (pi/2, pi)` But `theta ne pi/4` (not in domain), therefore, `theta in (pi/2, pi)` |
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| 230. |
The least positive solution of `cot (pi/(3 sqrt(3)) sin 2x)=sqrt(3)` lies inA. `(0, pi/6]`B. `(pi/9, pi/6)`C. `(pi/12, pi/9]`D. `(pi/3, pi/2]` |
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Answer» Correct Answer - A `cot (pi/(3sqrt(3)) sin 2 x)=sqrt(3)` `rArr pi/(3sqrt(3)) sin 2x=n pi +pi/6, n in Z` `rArr sin 2x=3sqrt(3) n + sqrt(3)/2` For solution, `n=0` `rArr sin 2x=sqrt(3)/2` `rArr 2x=pi/3 rArr x=pi/6` |
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| 231. |
For `0 |
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Answer» Correct Answer - C We have `cos y=sqrt(3)/2` `rArr y=pi/6` Let `cot (x-y)=t` `:. t^(2)-t-sqrt(3) t+sqrt(3)=0` `rArr (t-1) (t-sqrt(3))=0` `:. cot (x-y)=1 or cot (x-y)=sqrt(3)` Given that `x in (0, pi)` If `x-y=pi/6` then `x=pi/6+pi/6=pi/3 rArr (x, y) =(pi/3, pi/6)` If `x-y=pi/4` then `x=pi/6+pi/4=(5pi)/12 rArr (x, y) = ((5pi)/12, pi/6)` Possible ordered pairs (x, y) are `(pi/3, pi/6), ((5pi)/12, pi/6)`. |
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| 232. |
Find the general solution of the equation `3^(sin2x+2cos^2x)+3^(1-sin2x+2sin^2x)=28`A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - B `3^(sin 2x + 2cos^(2)x)+3^(1+2(1-cos^(2)x)-sin 2x)=28` `rArr 3^(sin 2 x + 2 cos^(2)x)+(27)/(3^(sin 2x + 2 cos^(2)x))=28` `rArr (3^(t))^(2)-28(3^(t))+27=0 rArr 3^(t)=27` or 1 `rArr 2x + 2 cos^(2)x=0` or 3 (not possible) `rArr sin 2x + 2 cos^(2) x =0` `rArr sin 2x + cos 2x =-1` `rarr 4x = n pi, n in Z` `rArr x = (3pi)/(4),(7pi)/(4),(pi)/(2),(3pi)/(2)` |
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| 233. |
Find the general solution of each of the equations : (i) `sin2x=-(1)/(2)` (ii) `tan3x=-1` |
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Answer» (i) `sin2x=-(1)/(2)=-"sin"(pi)/(6)=sin(pi+(pi)/(6))="sin "(7pi)/(6)` `rArrsin2x="sin"(7pi)/(6)` `rArr2x={npi+(-1)^(n)*(7pi)/(6)}` where `ninl` `rArrx={(npi)/(2)+(-1)^(n)*(7pi)/(12)}`, where `ninl`. Hence , the general solution is x `={(npi)/(2)+(-1)^(n)*(7pi)/(12)}` , where `ninl`. (ii) `tan3x=-1=-"tan"(pi)/(4)=tan(pi-(pi)/(4))="tan"(3pi)/(4)` `rArrtan3x" tan"(3pi)/(4)` `rArr3x=(npi+(3pi)/(4))` where `ninl`. `rArrx=((npi)/(3)+(pi)/(4))`, where `ninl`. Hence , the general solution is `x=((npi)/(3)+(pi)/(4))` where `ninl`. |
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