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Question

The sum of all roots of $sin invisible function application open parentheses pi log subscript 3 end subscript invisible function application open parentheses fraction numerator 1 over denominator x end fraction close parentheses close parentheses equals 0$ in $blank open parentheses 0 comma blank 2 pi close parentheses$ is

$3 / 2$
4
$9 / 2$
$13 / 3$

The correct answer is: $9 divided by 2$

$pi log subscript 3 end subscript invisible function application open parentheses fraction numerator 1 over denominator x end fraction close parentheses equals k pi comma blank k element of I$
$log subscript 3 end subscript invisible function application open parentheses fraction numerator 1 over denominator x end fraction close parentheses equals k rightwards double arrow x equals 3 to the power of negative k end exponent$
The possible values of $k$ are $negative 1 comma blank 0 comma blank 1 comma blank 2 comma blank 3 comma blank horizontal ellipsis$
$S equals 3 plus 1 plus fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator 3 to the power of 3 end exponent end fraction plus horizontal ellipsis infinity equals fraction numerator 3 over denominator 1 minus fraction numerator 1 over denominator 3 end fraction end fraction equals fraction numerator 9 over denominator 2 end fraction$

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If $sin invisible function application theta subscript 1 end subscript plus sin invisible function application theta subscript 2 end subscript plus sin invisible function application theta subscript 3 end subscript equals 3 comma$ then $cos invisible function application theta subscript 1 end subscript plus cos invisible function application theta subscript 2 end subscript plus cos invisible function application theta subscript 3 end subscript$ is equal to

If $sin invisible function application theta subscript 1 end subscript plus sin invisible function application theta subscript 2 end subscript plus sin invisible function application theta subscript 3 end subscript equals 3 comma$ then $cos invisible function application theta subscript 1 end subscript plus cos invisible function application theta subscript 2 end subscript plus cos invisible function application theta subscript 3 end subscript$ is equal to

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In any triangle $A B C comma fraction numerator a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent over denominator R to the power of 2 end exponent end fraction$ has the maximum value of

In any triangle $A B C comma fraction numerator a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent over denominator R to the power of 2 end exponent end fraction$ has the maximum value of

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In triangle $A B C comma blank a equals 5 comma blank b equals 4 blank$ and $blank c equals 3. blank G$ is the centroid of the triangle. Circumradius of triangle $G A B$ is equal to

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Number of ordered pair $open parentheses s close parentheses open parentheses a comma blank b close parentheses$ for each of which the equality $a open parentheses cos invisible function application x minus 1 close parentheses plus b to the power of 2 end exponent equals cos invisible function application open parentheses a x plus b to the power of 2 end exponent close parentheses minus 1$ holds true for all $x element of R$ are

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$I f tan invisible function application x equals b divided by a comma blank t h e n blank square root of open parentheses a plus b close parentheses divided by open parentheses a minus b close parentheses end root plus square root of open parentheses a minus b close parentheses divided by open parentheses a plus b close parentheses end root$ is equal to

$I f tan invisible function application x equals b divided by a comma blank t h e n blank square root of open parentheses a plus b close parentheses divided by open parentheses a minus b close parentheses end root plus square root of open parentheses a minus b close parentheses divided by open parentheses a plus b close parentheses end root$ is equal to

The total number of solution of $sin to the power of 4 end exponent invisible function application x plus cos to the power of 4 end exponent invisible function application x equals sin invisible function application x cos invisible function application x$ in $blank open square brackets 0 comma blank 2 pi close square brackets blank$ is equal to

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parallel

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