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Hint: Use the double angle formulae for $ \sin 2\theta =2\sin \theta \cos \theta $ and $ \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ to simply the equation and reduce it in terms of $ \theta $ . Further use algebraic operations to further reduce the equation in terms of only $ \sin \theta $ on the LHS and $ \cos \theta $ on the RHS. Factorise the equation obtained to get two solution sets. Finally, use the formula for general solutions of $ \sin \theta =\sin \phi $ to find the values of $ \theta $ .

Complete step-by-step answer:

We know that both the expressions on the RHS, presently in terms of $ 2\theta $ can be expressed as trigonometric ratios of $ \theta $ by using the formulae for $ \sin 2\theta $ and $ \cos 2\theta $ which are given as $ \sin 2\theta =2\sin \theta \cos \theta $ and $ \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ .

Using these formulae on the RHS, the equation thus becomes

$ \cos \theta +\sin \theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta +2\sin \theta \cos \theta $

We rearrange the terms such that all the terms having $ \sin \theta $ are on the LHS and all the terms having $ \cos \theta $ are on the RHS. Thus,

$ \sin \theta +{{\sin }^{2}}\theta ={{\cos }^{2}}\theta -\cos \theta +2\sin \theta \cos \theta $

Now we subtract $ \sin \theta \cos \theta $ from both sides of this equation to get the same term involving $ \sin \theta $ and $ \cos \theta $ on both sides of the equation, so as to get

\[\sin \theta +{{\sin }^{2}}\theta -\sin \theta \cos \theta ={{\cos }^{2}}\theta -\cos \theta +\sin \theta \cos \theta \]

In this equation, we take $ \sin \theta $ common on the LHS and $ \cos \theta $ common on the RHS. This gives us

\[\begin{align}

& \sin \theta \left( 1+\sin \theta -\cos \theta \right)=\cos \theta \left( 1+\sin \theta -\cos \theta \right) \\

& \Rightarrow \sin \theta \left( 1+\sin \theta -\cos \theta \right)-\cos \theta \left( 1+\sin \theta -\cos \theta \right)=0 \\

& \Rightarrow \left( \sin \theta -\cos \theta \right)\left( 1+\sin \theta -\cos \theta \right)=0 \\

\end{align}\]

Thus the possible solutions occur when either \[\sin \theta -\cos \theta =0\] or \[1+\sin \theta -\cos \theta =0\] .

Case I: When \[\sin \theta -\cos \theta =0\] .

$ \begin{align}

& \Rightarrow \sin \theta =\cos \theta \\

& \Rightarrow \sin \theta =\sin \left( {{90}^{\circ }}-\theta \right) \\

\end{align} $

A general solution of the equation $ \sin \theta =\sin \phi $ is given by $ \theta =n\pi +{{\left( -1 \right)}^{n}}\phi $ . Using this general solution for the above equation, we get

$ \theta =n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2}-\theta \right) $ .

The values can be found by using $ n=0,1,... $ to get

$ \theta =\dfrac{\pi }{4},\ \dfrac{5\pi }{4},\ \dfrac{9\pi }{4},\ \ldots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) $

Case II: When \[1+\sin \theta -\cos \theta =0\]

Use the identity $ \sin \theta -\cos \theta =\sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right) $ . Using this identity, the equation is resolved to

$ \begin{align}

& 1+\sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right)=0 \\

& \Rightarrow \sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right)=-1 \\

& \Rightarrow \sin \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{-1}{\sqrt{2}} \\

& \Rightarrow \sin \left( \theta -\dfrac{\pi }{4} \right)=\sin \dfrac{3\pi }{4} \\

\end{align} $

Finding the general solution by $ \theta =n\pi +{{\left( -1 \right)}^{n}}\phi $ where $ \theta =\theta -\dfrac{\pi }{4} $ and $ \phi =\dfrac{3\pi }{4} $ . Thus,

$ \theta -\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{3\pi }{4} $

Using values of $ n=0,1,\ldots $ the values of $ \theta $ can be obtained as

$ \theta =\dfrac{\pi }{2},\pi ,\dfrac{5\pi }{2},3\pi ,\dfrac{9\pi }{2},5\pi ,\ldots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) $

Combining the result from equations (1) and (2), we get

$ \theta =\dfrac{\pi }{4},\dfrac{\pi }{2},\pi ,\dfrac{5\pi }{4},\dfrac{9\pi }{4},\dfrac{5\pi }{2},3\pi ,\ldots $

Thus, these are the required values of $ \theta $ .

Note: The trick here is decomposing the given expression into a factorable expression. For this, aim at resolving into as simple terms of $ \theta $ as possible and try to eliminate complex terms by simple algebraic operations. Finally, it should be kept in mind that the solution sets of both the equations need to be considered and not just one. Also, general solutions need to be considered and particular solutions can be obtained from it by substituting the values of n.

Complete step-by-step answer:

We know that both the expressions on the RHS, presently in terms of $ 2\theta $ can be expressed as trigonometric ratios of $ \theta $ by using the formulae for $ \sin 2\theta $ and $ \cos 2\theta $ which are given as $ \sin 2\theta =2\sin \theta \cos \theta $ and $ \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ .

Using these formulae on the RHS, the equation thus becomes

$ \cos \theta +\sin \theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta +2\sin \theta \cos \theta $

We rearrange the terms such that all the terms having $ \sin \theta $ are on the LHS and all the terms having $ \cos \theta $ are on the RHS. Thus,

$ \sin \theta +{{\sin }^{2}}\theta ={{\cos }^{2}}\theta -\cos \theta +2\sin \theta \cos \theta $

Now we subtract $ \sin \theta \cos \theta $ from both sides of this equation to get the same term involving $ \sin \theta $ and $ \cos \theta $ on both sides of the equation, so as to get

\[\sin \theta +{{\sin }^{2}}\theta -\sin \theta \cos \theta ={{\cos }^{2}}\theta -\cos \theta +\sin \theta \cos \theta \]

In this equation, we take $ \sin \theta $ common on the LHS and $ \cos \theta $ common on the RHS. This gives us

\[\begin{align}

& \sin \theta \left( 1+\sin \theta -\cos \theta \right)=\cos \theta \left( 1+\sin \theta -\cos \theta \right) \\

& \Rightarrow \sin \theta \left( 1+\sin \theta -\cos \theta \right)-\cos \theta \left( 1+\sin \theta -\cos \theta \right)=0 \\

& \Rightarrow \left( \sin \theta -\cos \theta \right)\left( 1+\sin \theta -\cos \theta \right)=0 \\

\end{align}\]

Thus the possible solutions occur when either \[\sin \theta -\cos \theta =0\] or \[1+\sin \theta -\cos \theta =0\] .

Case I: When \[\sin \theta -\cos \theta =0\] .

$ \begin{align}

& \Rightarrow \sin \theta =\cos \theta \\

& \Rightarrow \sin \theta =\sin \left( {{90}^{\circ }}-\theta \right) \\

\end{align} $

A general solution of the equation $ \sin \theta =\sin \phi $ is given by $ \theta =n\pi +{{\left( -1 \right)}^{n}}\phi $ . Using this general solution for the above equation, we get

$ \theta =n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2}-\theta \right) $ .

The values can be found by using $ n=0,1,... $ to get

$ \theta =\dfrac{\pi }{4},\ \dfrac{5\pi }{4},\ \dfrac{9\pi }{4},\ \ldots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) $

Case II: When \[1+\sin \theta -\cos \theta =0\]

Use the identity $ \sin \theta -\cos \theta =\sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right) $ . Using this identity, the equation is resolved to

$ \begin{align}

& 1+\sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right)=0 \\

& \Rightarrow \sqrt{2}\sin \left( \theta -\dfrac{\pi }{4} \right)=-1 \\

& \Rightarrow \sin \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{-1}{\sqrt{2}} \\

& \Rightarrow \sin \left( \theta -\dfrac{\pi }{4} \right)=\sin \dfrac{3\pi }{4} \\

\end{align} $

Finding the general solution by $ \theta =n\pi +{{\left( -1 \right)}^{n}}\phi $ where $ \theta =\theta -\dfrac{\pi }{4} $ and $ \phi =\dfrac{3\pi }{4} $ . Thus,

$ \theta -\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{3\pi }{4} $

Using values of $ n=0,1,\ldots $ the values of $ \theta $ can be obtained as

$ \theta =\dfrac{\pi }{2},\pi ,\dfrac{5\pi }{2},3\pi ,\dfrac{9\pi }{2},5\pi ,\ldots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) $

Combining the result from equations (1) and (2), we get

$ \theta =\dfrac{\pi }{4},\dfrac{\pi }{2},\pi ,\dfrac{5\pi }{4},\dfrac{9\pi }{4},\dfrac{5\pi }{2},3\pi ,\ldots $

Thus, these are the required values of $ \theta $ .

Note: The trick here is decomposing the given expression into a factorable expression. For this, aim at resolving into as simple terms of $ \theta $ as possible and try to eliminate complex terms by simple algebraic operations. Finally, it should be kept in mind that the solution sets of both the equations need to be considered and not just one. Also, general solutions need to be considered and particular solutions can be obtained from it by substituting the values of n.

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