Question

The sum of all values of $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ satisfying ${{\sin }^{2}}2\theta +{{\cos }^{4}}2\theta =\dfrac{3}{4}$ is?
A. $\dfrac{\pi }{2}$
B. $\pi $
C. $\dfrac{3\pi }{8}$
D. $\dfrac{5\pi }{4}$

Answer 1

Hint: In the above question we will substitute the value of ${{\sin }^{2}}2\theta $ in form of ${{\cos }^{2}}2\theta $ by using the trigonometric identity as follows:
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
After that we will use the general solution for \[{{\cos }^{2}}\theta ={{\cos }^{2}}\alpha \] given by,
$\theta =n\pi \pm \alpha $

Complete step-by-step answer:

We have been asked to find the sum of all values of $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ satisfying ${{\sin }^{2}}2\theta +{{\cos }^{4}}2\theta =\dfrac{3}{4}$.
We know that ${{\sin }^{2}}2\theta +{{\cos }^{2}}2\theta =1$, which is a trigonometric identity.
${{\sin }^{2}}2\theta =1-{{\cos }^{2}}2\theta $
Substituting the values of ${{\sin }^{2}}2\theta $ in the given equation, we get,
$1-{{\cos }^{2}}2\theta +{{\cos }^{4}}2\theta =\dfrac{3}{4}$
On rearranging the terms, we get,
$\begin{align}
  & {{\left( {{\cos }^{2}}2\theta \right)}^{2}}-{{\cos }^{2}}2\theta +1-\dfrac{3}{4}=0 \\
 & {{\left( {{\cos }^{2}}2\theta \right)}^{2}}-{{\cos }^{2}}2\theta +\dfrac{1}{4}=0 \\
\end{align}$
The above equation is in the form of ${{x}^{2}}-2ax+{{a}^{2}}$ which is equal to ${{\left( x-a \right)}^{2}}$.
Here, $x={{\cos }^{2}}2\theta \ and\ a=\dfrac{1}{2}$.
$\begin{align}
  & \Rightarrow {{\left( {{\cos }^{2}}2\theta -\dfrac{1}{2} \right)}^{2}}=0 \\
 & \Rightarrow {{\cos }^{2}}2\theta =\dfrac{1}{2}={{\cos }^{2}}\left( \dfrac{\pi }{4} \right) \\
\end{align}$
Since, $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$.
We know the general solution for ${{\cos }^{2}}\theta ={{\cos }^{2}}\alpha $ is given by,
$\begin{align}
  & \theta =n\pi \pm \alpha \\
 & \Rightarrow 2\theta =n\pi \pm \dfrac{\pi }{4} \\
\end{align}$
For $n=0$,
$\begin{align}
  & 2\theta =0\pm \dfrac{\pi }{4} \\
 & \theta =\pm \dfrac{\pi }{8} \\
\end{align}$
Since, $\theta \in \left( 0,\dfrac{\pi }{2} \right)\Rightarrow \theta =\dfrac{\pi }{8}$
For $n=1$,
$\begin{align}
  & 2\theta =\pi \pm \dfrac{\pi }{4} \\
 & 2\theta =\pi +\dfrac{\pi }{4} \\
 & \Rightarrow \theta =\dfrac{5\pi }{8} \\
 & 2\theta =\pi -\dfrac{\pi }{4} \\
 & \Rightarrow \theta =\dfrac{3\pi }{8} \\
\end{align}$
Since, $\dfrac{5\pi }{8}>\dfrac{\pi }{2}$. So, we cannot take this value according to the given condition in the question.
$\Rightarrow \theta =\dfrac{3\pi }{8}$ satisfies the question.
Hence, the solutions for the given equation are $\dfrac{\pi }{8}\ and\ \dfrac{3\pi }{8}$.
So, their sum $=\dfrac{\pi }{8}+\dfrac{3\pi }{8}$
 $\begin{align}
  & =\dfrac{\pi +3\pi }{8} \\
 & =\dfrac{4\pi }{8} \\
 & =\dfrac{\pi }{2} \\
\end{align}$
Therefore, the correct option is (A).

Note: Be careful while choosing the option as sometimes we just forget that we have to find the sum of the values that satisfy the equation instead of it we choose the value i.e. option (C) $\dfrac{3\pi }{8}$.
Also, remember that while finding the value of $'\theta '$ take care of the given condition on $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ otherwise you will get an extra root.