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Maths-

General

Easy

Question

$I f cos invisible function application x plus cos invisible function application y minus cos invisible function application open parentheses x plus y close parentheses equals fraction numerator 3 over denominator 2 end fraction comma blank t h e n$

$x + y = 0$
$x = 2 y$
$x = y$
$2 x = y$

The correct answer is: $x equals y$

$cos invisible function application x plus cos invisible function application y minus cos invisible function application open parentheses x plus y close parentheses equals fraction numerator 3 over denominator 2 end fraction$
$rightwards double arrow 2 cos invisible function application open parentheses fraction numerator x plus y over denominator 2 end fraction close parentheses cos invisible function application open parentheses fraction numerator x minus y over denominator 2 end fraction close parentheses minus 2 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator x plus y over denominator 2 end fraction close parentheses plus 1 equals fraction numerator 3 over denominator 2 end fraction$
$rightwards double arrow 2 cos invisible function application blank to the power of 2 end exponent open parentheses fraction numerator x plus y over denominator 2 end fraction close parentheses minus 2 cos invisible function application open parentheses fraction numerator x plus y over denominator 2 end fraction close parentheses cos invisible function application open parentheses fraction numerator x minus y over denominator 2 end fraction close parentheses plus fraction numerator 1 over denominator 2 end fraction equals 0$
Now $cos invisible function application open parentheses fraction numerator x plus y over denominator 2 end fraction close parentheses$ is always real, then discriminant $greater or equal than 0$
$rightwards double arrow 4 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator x minus y over denominator 2 end fraction close parentheses minus 4 greater or equal than 0$
$rightwards double arrow cos to the power of 2 end exponent invisible function application open parentheses fraction numerator x minus y over denominator 2 end fraction close parentheses greater or equal than 1$
$rightwards double arrow cos to the power of 2 end exponent invisible function application open parentheses fraction numerator x minus y over denominator 2 end fraction close parentheses equals 1$
$rightwards double arrow fraction numerator x minus y over denominator 2 end fraction equals 0 rightwards double arrow x equals y$

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