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`3//2`4`9//2``13//3`

Answer :

CSolution :

`pi log_(3) (1/x)=k pi, k in I` <br> `log_(3) (1/x)=k or x=3^(-k)` <br> The possible values of k are `-1, 0, 1, 2, 3,...` <br> `S=3+1+1/3+1/3^(2)+1/3^(3)+...oo=3/(1-1/3)=9/2` Transcript

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00:00 - 00:59 | hello this is the given question and this question we have the sum of all the roots of equation that is sin file or three one upon X is equals to zero this equation given to us at state that we have to find the sum of all the roots of the equation in limit 0 to 2pi so options are given that is 3 by 2 option 1 option 20 for efficiency is 9 by 2013 by three ok festival will write the situation that will try to solve it ok so this is the given equation here you can see the angle which sign is this ok so lets considered this ok let's represent this angle with something so that the calibration will be easier so let's take this is equals to let me right it by log 3 1 upon X ok let's |

01:00 - 01:59 | ok now you can see we have just consider this Lok 3 1 upon x as ke and we have added by because here it was why already present ok so we could also have just returned let me just to you could also have done this so basically what I want to say our value for ke is equals to log off 3 1 upon X ok so from here you will find the value of x in order to find the value of x what will do ok log 3 1 upon X means X raise to power minus 1 is equal to ke ok simple so if anything is written power with lock it comes in forward it is a rule so we can write |

02:00 - 02:59 | this medicine coming forward we can rewrite this as - of log 3 X is equals to ke or we can write log 3 log of 3 X is equals to minus ke ok simple now what will do will find the value of x from here ok so let's should this lock to Rs whenever we ship to log to Other Side then it becomes this whatever is written here if suppose if log x equals to 3 then X is equals to this base we write here and in Rs we have we write it in our ok this is acceleration but this part is just explanation but basically what I want to say here we have three years bass |

03:00 - 03:59 | here in this equation vs DS base so X will be free and raise to power minus ke ok simple property of log Sabhi Karte value of x that is this ok so now let me that the given equation the question again then I'll try to make you understand ok so this was the question given to those in the question now when we look at the situation we have where one upon X ok so this equation will be satisfied only if x value X value will be either 3 or multiple of 3 here we have 3 - ke so I will put any other value this equation that is given to us in the question will not satisfied ok mein X should be either 3 or multiple of 3 ok or I can also say look sign if this whole value will get zero we know |

04:00 - 04:59 | 10000 it is also satisfied for sin 0 7 0 is equal to zero it is also satisfied for this and it is also satisfied for multiple ok multiples of 3 ok so this is equation is satisfied for these two conditions and look how I can get here in this equation in this equation how I can get zero here if I will get zero here if my ex will be if my actual B1 look how I can get X equal to 1 if I'll put you in place of minus ke then let me show you if ke Kuchh To zero then X is equals to 3 raise to power 0 it means X = 21 ok and we know that it will be |

05:00 - 05:59 | and value of log10 suicide 0 will be zero basically what I want to say my case value will be there for I want to say that ke will be either -1 if I'll put minus one place of ke xb3 OK it is a multiple of 3 condition satisfied if I'll put 0 already show it here ok if I put one so it will be three days by minus one that is X will be 1 upon 3 so I again got multiple of 3 similarly for 237 ok Sudheer cash value and if I'll put ke if I'll put ke hair but I'll get I have to find the sum of all the roots OS is equals to file Put Ke call to minus 1 then axle B3 as I told yourself with three plus because I have to |

06:00 - 06:59 | some people Put Ke = 20 then I'll get one I told you here I get one ok if I'll put equal to 1 then I'll get 1 upon 3 OK it is also a multiple of 3 if I'll put ke equals to I'll get one upon 9 ok similarly this will go to infinity now I have to find the sum so some will be = 27 will be equal to 8 that is first number divided by 1 minus 1 upon 3 difference between these two numbers ok we know that how to find sum to infinity first number that is this and difference between these numbers ok so on taking LCM I'll get here 3 upon 3 minus one and this 3 will come in numerator it will be 9 upon |

07:00 - 07:59 | show my Sangh sum of all the roots of the equation is 9 b to a sum is 9 Main To so let's check ok options she is it is clearly visible high efficiency is meant by two so I have some is 9 Main To so I can clearly say that option the options she is it ok I can clearly claim that option C is correct option among all the four option thank you |

**Trigonometric equations and their solutions**

**(i) Prove that general solution of `sintheta=0` is given by `theta=npi; n in Z` (ii) Prove that general solution of `costheta=0` is `theta=((2n+1)pi/2); n in Z`**

**(iii)Prove that general solution of `tan theta=0` is `theta=npi; n in Z` (iv)Prove that the general solution of `cot theta=0` is `theta=(2n+1)pi/2, n in Z`**

**Prove that general solution of `sintheta=sin alpha` is given by `theta=npi+(-1)^n alpha, n in Z`**

**Find the general solution of the equations (i)`sin theta= sqrt3/2` (ii) `2sintheta+1=0` (iii)`cosectheta=2`**

**Prove that the general solution of `costheta=cosalpha`is given by `theta=2npipmalpha`; where `n inZ`**

**Find the general solution of the equation (i)`costheta=1/2 (ii) cos3theta=-1/2 (iii)sqrt3sec2theta=2`**

**Prove that general solution of `tan theta= tan alpha` is given by `theta=npi+alpha;n in Z`**

**Find the general solution of (i) `tan theta=1/sqrt3` (ii) `tan2theta=sqrt3` (iii)`tan3theta=-1`**

**General solution of `(i) sin^2theta = sin^2alpha; (ii) cos^2 theta = cos^2 alpha (iii) tan^2 theta = tan^2 alpha`**