The equation $ {{3}^{\sin 2x+2{{\left( \cos x \right)}^{2}}}}+{{3}^{1-\sin 2x+2{{\left( \sin x \right)}^{2}}}}=28 $ is satisfied for the values of x given by
A. $ \dfrac{3\pi }{8} $
B. $ \dfrac{3\pi }{2} $
C. $ \dfrac{3\pi }{4} $
D. $ \dfrac{\pi }{8} $

Hint: In this question we have exponential in L.H.S side and R.H.S have natural number, seems to be complex to solve. Then first we will make the power with the same angle of trigonometry, as in our case it is ‘x’ and ‘2x’. So to make it simple first we will make angles equal. Which on further solving gives us a simpler equation, from where we will get the values of the assumed exponential from there we can calculate the value of ‘x’ which will satisfy the equation.

Complete step by step answer:
Moving further with the question, we have $ {{3}^{\sin 2x+2{{\left( \cos x \right)}^{2}}}}+{{3}^{1-\sin 2x+2{{\left( \sin x \right)}^{2}}}}=28 $ ; first we will simplify to make trigonometric angle equal, that in; $ \sin 2x+2{{\left( \cos x \right)}^{2}} $ and $ 1-\sin 2x+2{{\left( \sin x \right)}^{2}} $
So; as we known that
 $ \begin{align}
  & 2{{\left( \cos x \right)}^{2}}-1=\cos 2x \\
 & 2{{\left( \cos x \right)}^{2}}=1+\cos 2x \\
\end{align} $
Is a trigonometric identity, so we can $ \sin 2x+2{{\left( \cos x \right)}^{2}} $ (which is exponent of 3) as;
 $ \begin{align}
  & \sin 2x+2{{\left( \cos x \right)}^{2}} \\
 & \sin 2x+1+\cos 2x \\
 & \sin 2x+\cos 2x+1 \\
\end{align} $ equation (i)
And, similarly we also know that;
 $ \begin{align}
  & 1-2{{\left( \sin x \right)}^{2}}=\cos 2x \\
 & 1-\cos 2x=2{{\left( \sin x \right)}^{2}} \\
\end{align} $
From the trigonometric identity, so we can write $ 1-\sin 2x+2{{\left( \sin x \right)}^{2}} $ (which is exponent of 3) as;
 $ \begin{align}
  & 1-\sin 2x+2{{\left( \sin x \right)}^{2}} \\
 & 1-\sin 2x+1-\cos 2x \\
 & 2-(\sin 2x+\cos 2x) \\
\end{align} $ equation (ii)
So from equation (i) and (ii) replace $ \sin 2x+2{{\left( \cos x \right)}^{2}} $ and $ 1-\sin 2x+2{{\left( \sin x \right)}^{2}} $ with $ \sin 2x+\cos 2x+1 $ and $ 2-(\sin 2x+\cos 2x) $ respectively from the given equation in question, which on further simplifying we can write it as;
 \[\begin{align}
  & {{3}^{\sin 2x+2{{\left( \cos x \right)}^{2}}}}+{{3}^{1-\sin 2x+2{{\left( \sin x \right)}^{2}}}}=28 \\
 & {{3}^{1+\sin 2x+\cos 2x}}+{{3}^{2-\left( \sin 2x+\cos 2x \right)}}=28 \\
 & {{3}^{1}}{{.3}^{\sin 2x+\cos 2x}}+\dfrac{{{3}^{2}}}{{{3}^{\sin 2x+\cos 2x}}}=28 \\
\end{align}\]
Since in this equation \[{{3}^{\sin 2x+\cos 2x}}\] is common, let us assume it to be equal to ‘a’. So we will have new equation that is;
 \[{{3}^{1}}.a+\dfrac{{{3}^{2}}}{a}=28\]
By taking the LCM which on solving it further, bringing all values to one side, we will get;
 \[\begin{align}
  & 3{{a}^{2}}+{{3}^{2}}=28a \\
 & 3{{a}^{2}}+28a+9=0 \\
\end{align}\]
Now, to find the value of ‘a’ by middle term splitting, we will get;
 \[\begin{align}
  & 3{{a}^{2}}-28a+9=0 \\
 & 3{{a}^{2}}-27a-a+9=0 \\
 & 3a(a-9)-(a-9)=0 \\
 & (a-9)(3a-1)=0 \\
\end{align}\]
So from here we can say that \[(a-9)=0\] or \[(3a-1)=0\]
 \[\begin{align}
  & a-9=0 \\
 & a=9 \\
 & or \\
 & 3a-1=0 \\
 & a=\dfrac{1}{3} \\
\end{align}\]
From here we can say that \[a=9\] or \[a=\dfrac{1}{3}\] , as ‘a’ is the assumed variable of \[{{3}^{\sin 2x+\cos 2x}}\] . So we can further write it as;
 \[\begin{align}
  & {{3}^{\sin 2x+\cos 2x}}=9 \\
 & {{3}^{\sin 2x+\cos 2x}}={{3}^{2}} \\
 & \sin 2x+\cos 2x=2 \\
\end{align}\]
As we know that \[\sin x+\cos x\] lies between $ \left( -\sqrt{2},\sqrt{2} \right) $ , whatever may be angle, so in our case \[\sin 2x+\cos 2x=2\] is never possible; because $ \sin 2x+\cos 2x $ always lie between $ \left( -\sqrt{2},\sqrt{2} \right) $ and it says equal to 2. So this option is not correct. Let us now check for other condition i.e. \[a=\dfrac{1}{3}\] , so we can write it as;
 \[\begin{align}
  & {{3}^{\sin 2x+\cos 2x}}=\dfrac{1}{3} \\
 & {{3}^{\sin 2x+\cos 2x}}={{3}^{-1}} \\
 & \sin 2x+\cos 2x=-1 \\
\end{align}\]
This case can be possible, as $ -1 $ lies between $ \left( -\sqrt{2},\sqrt{2} \right) $
As there can be many values of x that will satisfy the condition, so for simplicity we can now put the value of x to satisfy the condition \[\sin 2x+\cos 2x=-1\] , and whichever case will satisfy that will be the answer.
So correct answer will be option ‘B’ and ‘’ i.e. $ \dfrac{3\pi }{2} $ and $ \dfrac{3\pi }{4} $

Note: Always remember that \[\sin x+\cos x\] will have range of $ \left( -\sqrt{2},\sqrt{2} \right) $ , whatever might be the angle, the function can never be greater than $ \left( -\sqrt{2},\sqrt{2} \right) $ . Moreover, to find the exact solution rather than going with options as we did in the last, we can go with the concept of trigonometric equations of finding general solutions.