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Question

e general solution of the equation $8 cos invisible function application x cos invisible function application 2 x cos invisible function application 4 x equals sin invisible function application 6 x divided by sin invisible function application x$ is

$x = (n π / 7) + (π / 21), \forall n \in Z$
$x = (2 π / 7) + (π / 14), \forall n \in Z$
$x = (n π / 7) + (π / 14), \forall n \in Z$
$x = (n π) + (π / 14), \forall n \in Z$

The correct answer is: $x equals open parentheses n pi divided by 7 close parentheses plus open parentheses pi divided by 14 close parentheses comma blank for all n element of Z$

The given equation is $8 sin invisible function application x cos invisible function application x cos invisible function application 2 x cos invisible function application 4 x equals sin invisible function application 6 x open parentheses sin invisible function application x not equal to 0 close parentheses$
$rightwards double arrow sin invisible function application 8 x equals sin invisible function application 6 x blank rightwards double arrow 2 cos invisible function application 7 x sin invisible function application x equals 0$
As $sin invisible function application x not equal to 0 comma cos invisible function application 7 x equals 0$ or $blank 7 x equals n pi plus pi divided by 2 comma blank n element of Z$
$i. e. comma blank x equals n pi divided by 7 plus pi divided by 14 semicolon n element of Z$

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parallel

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Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

$i s blank e q u a l blank t o$

$i s blank e q u a l blank t o$

parallel

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