Solve: ${sin}^{2} 60^{0} + {cos}^{2} (3 x - 9^{0}) = 1$

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Solution

To solve: ${sin}^{2} 60^{\circ} + {cos}^{2} (3 x - 9^{\circ}) = 1$

${sin}^{2} 60^{\circ} + {cos}^{2} (3 x - 9^{\circ}) = 1$
$\Rightarrow {cos}^{2} (3 x - 9^{\circ}) = 1 - {sin}^{2} 60^{\circ}$
$\Rightarrow {cos}^{2} (3 x - 9^{\circ}) = {cos}^{2} 60^{\circ}$ (Since, ${sin}^{2} θ + {cos}^{2} θ = 1$ )
$\Rightarrow 3 x - 9^{\circ} = 60^{\circ}$
$\Rightarrow 3 x = 69^{\circ}$
$\Rightarrow x = 23^{\circ}$

Hence, the value of $x$ is $23^{\circ}$

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