Question

If $sin3\theta =cos(\theta -6^o)$ then find the value of $\theta$ where $\left(3\theta \right)$ and $(\theta -6^o)$ are acute angles

Answer 1

$Sin3\theta =Cos(\theta -6^{\circ })$
$Cos(90-3\theta )=Cos(\theta -6^{\circ })$
${90}^0-3\theta =\theta -6$
$96=4\theta$
$\theta ={24}^{\circ }$