Question
Solve for $x : \sin^2 60^\circ + \cos^2 (3x-9^\circ ) = 1$
✓
Answer
$\sin^260^\circ + \cos^2(3x – 9^\circ ) = 1$
$\cos^2(3x – 9^\circ ) = 1 – \sin^260$
$\cos ^2\left(3 x-9^{\circ}\right)=1-\frac{3}{4}$
$\cos ^2\left(3 x-9^{\circ}\right)=\frac{1}{4}$
$\cos ^2\left(3 x-9^{\circ}\right)=\frac{1}{2}$
$3x – 9^\circ = 60^\circ$
$3x = 69^\circ$
$x = 23^\circ$
$\cos^2(3x – 9^\circ ) = 1 – \sin^260$
$\cos ^2\left(3 x-9^{\circ}\right)=1-\frac{3}{4}$
$\cos ^2\left(3 x-9^{\circ}\right)=\frac{1}{4}$
$\cos ^2\left(3 x-9^{\circ}\right)=\frac{1}{2}$
$3x – 9^\circ = 60^\circ$
$3x = 69^\circ$
$x = 23^\circ$
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