Question

$si{n}^2{60}^{\circ }+co{s}^2(3x-9^{\circ })=1$

Answer 1

$si{n}^2{60}^{\circ }+co{s}^2(3x-9^{\circ })=1$
$(\frac{\sqrt{3}}{2}{)}^2+co{s}^2(3x-9^{\circ })=1$
$co{s}^2(3x-9^{\circ })=1-\frac{3}{4}$
$co{s}^2(3x-9^{\circ })=\frac{1}{4}$
$cos(3x-9^{\circ })=\frac{1}{2}$
$cos(3x-9^{\circ })=cos60$
$3x-9=60$
$3x=69$
$x={23}^{\circ }$