Question

If \[0\le x\le \pi \] and \[{{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30\] then $x=$
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{3\pi }{4}$

Answer 1

Hint: To find the value of $x$ we will simplify the given equation using formula $1-{{\sin }^{2}}x={{\cos }^{2}}x$ and property ${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$. Using assumption method and then simplification we will derive a quadratic equation from where we will get two factors. Equating both factors to zero we will find all the possible values of $x$. Then we will select the value of $x$ in the interval \[0\le x\le \pi \].
The function sin is positive in second and third quadrant depicted as $\dfrac{\pi }{2}-\theta $ and $\dfrac{\pi }{2}+\theta ,\pi -\theta $.

Formula Used: $1-{{\sin }^{2}}x={{\cos }^{2}}x$

Complete step by step solution: We are given a trigonometric equation \[{{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30\] when \[0\le x\le \pi \] and we have to derive the value of $x$.
We will take the given equation and substitute the formula $1-{{\sin }^{2}}x={{\cos }^{2}}x$ in it.
\[{{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30\]
We will use the property ${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$ in the equation,
\[{{81}^{{{\sin }^{2}}x}}+\dfrac{{{81}^{1}}}{{{81}^{{{\sin }^{2}}x}}}=30\]
Let us assume ${{81}^{{{\sin }^{2}}x}}=z$.
Now the equation will be,
\[\begin{align}
  & z+\dfrac{81}{z}=30 \\
 & {{z}^{2}}+81=30z \\
 & {{z}^{2}}-30z+81=0
\end{align}\]
We will now factorize the derived quadratic equation.
\[\begin{align}
  & {{z}^{2}}-30z+81=0 \\
 & {{z}^{2}}-27z-3z+81=0 \\
 & z(z-27)-3(z-27)=0 \\
 & (z-27)(z-3)=0
\end{align}\]
Equating both factors to zero we will get $z=27,3$.
Now we will put back the value of $z$.
$\begin{align}
  & {{81}^{{{\sin }^{2}}x}}=27 \\
 & {{\left( {{\left( 3 \right)}^{4}} \right)}^{{{\sin }^{2}}x}}={{\left( 3 \right)}^{3}} \\
 & {{\left( 3 \right)}^{4{{\sin }^{2}}x}}={{\left( 3 \right)}^{3}} \\
 & 4{{\sin }^{2}}x=3 \\
 & {{\sin }^{2}}x=\dfrac{3}{4} \\
 & \sin x=\pm \dfrac{\sqrt{3}}{2}
\end{align}$ or $\begin{align}
  & {{81}^{{{\sin }^{2}}x}}=3 \\
 & {{\left( {{\left( 3 \right)}^{4}} \right)}^{{{\sin }^{2}}x}}={{\left( 3 \right)}^{1}} \\
 & {{\left( 3 \right)}^{4{{\sin }^{2}}x}}={{\left( 3 \right)}^{1}} \\
 & 4{{\sin }^{2}}x=1 \\
 & {{\sin }^{2}}x=\dfrac{1}{4} \\
 & \sin x=\pm \dfrac{1}{2}
\end{align}$
As the interval given for $x$ is \[0\le x\le \pi \] which falls in first and second quadrant where the function sin is positive . Now we know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. So,
$\sin x=\sin \dfrac{\pi }{6}$ or $\sin x=\sin \dfrac{\pi }{3}$ or $\sin x=-\sin \dfrac{\pi }{6}$ or $\sin x=-\sin \dfrac{\pi }{3}$
The possible values of $x$ are : $x=\dfrac{\pi }{6},\dfrac{\pi }{3},\pi -\dfrac{\pi }{3}=\dfrac{2\pi }{3},\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$.
The only answer present in the given options from the possible solutions of $x$ is $x=\dfrac{\pi }{4}$

Option ‘C’ is correct

Note: Instead of changing cos into sin we can also change the function sin into cos and then derive the value of $x$.