If ${sin}^{2} 60^{o} + {cos}^{2} (3 x - 9^{o}) = 1$ , then what is the value of $x$ ?

24^{o}

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23^{o}

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22^{o}

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21^{o}

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Solution

The correct option is B $23^{o}$
$s i n^{2} 60^{\circ} + c o s^{2} (3 x - 9^{\circ}) = 1$
$\Rightarrow c o s^{2} (3 x - 9^{\circ}) = 1 - \frac{3}{4} = \frac{1}{4}$
$\Rightarrow c o s (3 x - 9^{\circ}) = \frac{1}{2} = c o s (60^{\circ})$
$\Rightarrow 3 x - 9^{\circ} = 60^{\circ} \Rightarrow x = 23$

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