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# If $\sin 3\theta = \cos \left( {\theta - {6^0}} \right)$ then find the value of $\theta$ where $\left( {3\theta } \right)$ and $\left( {\theta - {6^0}} \right)$ are acute angles.  Hint- In order to find the value of a given angle with some condition. First in the problem statement we will use the trigonometric identities to convert the sine terms into cosine term or cosine term into sine term to make both the sides comparable.

Given that: $\sin 3\theta = \cos \left( {\theta - {6^0}} \right)$ ----- (1)
We will convert the sine term into the cosine term.
As we know that:
$\cos A = \sin \left( {{{90}^0} - A} \right)$
We will substitute the value of A as $\left( {\theta - {6^0}} \right)$ , so we have:
$\cos \left( {\theta - {6^0}} \right) \\ = \sin \left( {{{90}^0} - \left( {\theta - {6^0}} \right)} \right) \\ = \sin \left( {{{90}^0} - \theta + {6^0}} \right) \\ = \sin \left( {{{96}^0} - \theta } \right) \\$
Now let us put this value in equation (1)
$\because \sin 3\theta = \cos \left( {\theta - {6^0}} \right) \\ \Rightarrow \sin 3\theta = \sin \left( {{{96}^0} - \theta } \right) \\$
As we know that when $\sin A = \sin B$ then A = B when A and B are acute angles.
So using the same property in the above problem we get:
$\because \sin 3\theta = \cos \left( {\theta - {6^0}} \right) \\ \Rightarrow 3\theta = \left( {{{96}^0} - \theta } \right) \\ \Rightarrow 3\theta + \theta = {96^0} \\ \Rightarrow \theta = {24^0} \\$
Hence, the value of $\theta$ is ${24^0}$ for the given problem statement.

Note- In order to solve such types of problems students must remember the trigonometric identities relating one trigonometric term to another with change in angle. Also students may sometimes manipulate the identity according to the problem statement in order to bring the terms similar to the problem statement.
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