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Question

The total number of solutions of $open vertical bar cot invisible function application x close vertical bar equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction comma blank x element of open square brackets 0 comma blank 3 pi close square brackets$ is equal to

1
2
3
0

The correct answer is: 3

$open vertical bar cot invisible function application x close vertical bar equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction$
If $cot invisible function application x greater than 0 rightwards double arrow cot invisible function application x equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction equals 0$
$rightwards double arrow fraction numerator 1 over denominator sin invisible function application x end fraction equals 0 comma blank$ which is not possible
If $cot invisible function application x less or equal than 0 blank rightwards double arrow negative cot invisible function application x equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction$
$rightwards double arrow negative 2 cot invisible function application x equals fraction numerator 1 over denominator sin invisible function application x end fraction$
$rightwards double arrow cos invisible function application x equals negative fraction numerator 1 over denominator 2 end fraction$
$rightwards double arrow x equals fraction numerator 2 pi over denominator 3 end fraction comma fraction numerator 4 pi over denominator 3 end fraction comma fraction numerator 8 pi over denominator 3 end fraction blank open parentheses because x element of open square brackets 0 comma blank 3 pi close square brackets close parentheses$

Related Questions to study

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If the inequality $sin to the power of 2 end exponent invisible function application x plus a cos invisible function application x plus a to the power of 2 end exponent greater than 1 plus cos invisible function application x$ holds for any $x element of R comma$ then the largest negative integral value of $a$ is

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If $a to the power of 4 end exponent blank. b to the power of 5 end exponent equals 1 comma$ then the value of $log subscript a end subscript invisible function application open parentheses a to the power of 5 end exponent b to the power of 4 end exponent close parentheses$ equals

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If $open parentheses x plus 1 close parentheses to the power of log subscript 10 end subscript invisible function application open parentheses x plus 1 close parentheses end exponent equals 100 open parentheses x plus 1 close parentheses comma$ then

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The number of solutions of the equation $sin to the power of 3 end exponent invisible function application x cos invisible function application x plus sin to the power of 2 end exponent invisible function application x cos to the power of 2 end exponent invisible function application x plus sin invisible function application x cos to the power of 3 end exponent invisible function application x equals 1 comma blank$ in the interval $open square brackets 0 comma blank 2 pi close square brackets comma blank$ is

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parallel

If $a comma blank b blank$ and $blank c$ are the sides of a triangle, then the minimum value of $fraction numerator 2 a over denominator b plus c minus a end fraction plus fraction numerator 2 b over denominator c plus a minus b end fraction plus fraction numerator 2 c over denominator a plus b minus c end fraction$ is

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parallel

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parallel

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parallel

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