Let f(x)=(cosec^(4)x-2cosec^(2)x+1)/(cosec x(cosec x-sin x)+(si-Turito

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Let $f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction$ The sum of all the solutions of f (x) = 0 in [0, 100p] is

2550 $π$
2500 $π$
5000 $π$
5050 $π$

The correct answer is: 5000 $straight pi$

$table row cell table row blank cell f left parenthesis x right parenthesis equals fraction numerator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent over denominator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses plus 1 minus c o t invisible function application x plus c o t invisible function application x end fraction end cell row blank cell text defined for end text R minus n pi comma n element of I end cell row blank cell f left parenthesis x right parenthesis equals fraction numerator left parenthesis c o t invisible function application x right parenthesis to the power of 4 end exponent over denominator 1 plus c o t to the power of 2 end exponent invisible function application x end fraction end cell row blank cell f left parenthesis x right parenthesis equals 0 rightwards double arrow c o t invisible function application x equals 0 rightwards double arrow x equals left parenthesis 2 n minus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell row blank cell therefore x equals fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction comma fraction numerator 5 pi over denominator 2 end fraction comma horizontal ellipsis horizontal ellipsis comma fraction numerator 199 pi over denominator 2 end fraction end cell row blank cell s u m equals fraction numerator pi over denominator 2 end fraction left square bracket 1 plus 3 plus 5 plus horizontal ellipsis horizontal ellipsis plus 199 right square bracket end cell end table end cell row cell text end text text ( end text text 100 end text text end text text s end text text o end text text l end text text u end text text t end text text i end text text o end text text n end text text s end text text ) end text text end text end cell row cell equals fraction numerator pi over denominator 2 end fraction times fraction numerator 100 over denominator 2 end fraction times 200 equals 5000 pi end cell end table$

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Related Questions to study

The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n $element of$ I

We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x

x equals fraction numerator n pi over denominator 3 end fraction comma n element of I

The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n $element of$ I

We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x

x equals fraction numerator n pi over denominator 3 end fraction comma n element of I

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For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0

table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table

Then the sum of roots is 3p/2

The sum of all the solution of cotθ= sin 2θ(θ ≠ n $straight pi$ , n integer), 0 , $less or equal than theta less or equal than pi$ is

For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0

table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table

Then the sum of roots is 3p/2

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