Question

${ \sin }^{2} 60 + { \cos }^{2} (3x - 9) = 1$
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Answer 1

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Answer 2

$\fbox\blue{AnswEr:-}$

$Given \: sin^260° + cos {}^{2} (3x - 9°) = 1$

$= > ( \frac{ \sqrt{3} }{2} ) {}^{2} + cos {}^{2} (3x - 9) = 1$

$= > \frac{3}{4} + cos {}^{2} (3x - 9) = 1$

$= > cos {}^{2} (3x - 9) = 1 - \frac{3}{4}$

$= > cos {}^{2} (3x - 9) = \frac{1}{4}$

$= > cos(3x - 9) = \frac{ \sqrt{1} }{4}$

$= > cos(3x - 9) = \frac{1}{2}$

$∴ cos (3x-9)°= cos60° [∵ \frac{1}{2} = cos60°]$

$=>3x-9°=60°=>3x \: = 60°+9°$

$=>3x = 69° =>x = \frac{69°}{3}$

$\huge\pink{i.e. \: \: \: \: x = 23°}$