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Maths-

General

Easy

Question

Which of the following is not the general solution of $2 to the power of cos invisible function application 2 x end exponent plus 1 equals 3.2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent ?$

$n π, n \in Z$
$(n + \frac{1}{2}) π, n \in Z$
$(n - \frac{1}{2}) π, n \in Z$
None of these

The correct answer is: None of these

$cos invisible function application 2 x equals 1 minus 2 sin to the power of 2 end exponent invisible function application x blank$ and put $2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent equals t$
$rightwards double arrow 2 to the power of cos invisible function application 2 x end exponent equals 2 to the power of 1 minus 2 sin to the power of 2 end exponent invisible function application x end exponent equals 2 open parentheses 2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent close parentheses to the power of 2 end exponent equals 2 t to the power of 2 end exponent$
$rightwards double arrow 2 t to the power of 2 end exponent minus 3 t plus 1 equals 0$
$rightwards double arrow t equals 1 comma blank 1 divided by 2$
$rightwards double arrow 2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent equals 1 equals 2 to the power of 0 end exponent$
$rightwards double arrow sin to the power of 2 end exponent invisible function application x equals 0 blank$ or $blank x equals n pi comma blank n element of Z$
From $blank 2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent equals fraction numerator 1 over denominator 2 end fraction equals 2 to the power of negative 1 end exponent$ , we get
$sin to the power of 2 end exponent invisible function application x equals 1 blank o r blank x equals n pi plus-or-minus fraction numerator pi over denominator 2 end fraction comma blank n element of Z$

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