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Maths-

General

Easy

Question

The solution of $4 sin to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x plus c o s e c to the power of 2 end exponent x plus cot to the power of 2 end exponent invisible function application x minus 6 equals 0$ is

$n π \pm \frac{π}{4}$
$2 n π \pm \frac{π}{4}$
$n π + \frac{π}{3}$
$n π - \frac{π}{6}$

The correct answer is: $n pi plus-or-minus fraction numerator pi over denominator 4 end fraction$

$open parentheses 2 sin invisible function application x minus c o s e c blank x close parentheses to the power of 2 end exponent plus open parentheses tan invisible function application x minus cot invisible function application x close parentheses to the power of 2 end exponent equals 0$
$rightwards double arrow sin to the power of 2 end exponent invisible function application x equals fraction numerator 1 over denominator 2 end fraction blank a n d tan to the power of 2 end exponent invisible function application x equals 1$
$rightwards double arrow x equals n pi plus-or-minus fraction numerator pi over denominator 4 end fraction comma n element of Z$

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Maths-General

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