# If \[0\le x\le \pi \] and ${{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30$, then $x$ is equal

to.

(a) $\dfrac{\pi }{6}$

(b) $\dfrac{\pi }{3}$

(c) $\dfrac{5\pi }{6}$

(d) $\dfrac{2\pi }{3}$

(e) All correct

Last updated date: 22nd Mar 2023

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Answer

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Since the question contains both ${{\sin }^{2}}x$ and ${{\cos }^{2}}x$ , we will write ${{\cos}^{2}}x=1-{{\sin }^{2}}x$ so that we can get ${{81}^{{{\sin }^{2}}x}}$ in both the terms in left hand side of the equation given in the question. Then substitute ${{81}^{{{\sin }^{2}}x}}=t$ and solve the obtained quadratic equation.

In the question, we are given an equation ${{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos}^{2}}x}}=30.........(1)$.

We have a trigonometric identity,

$\begin{align}

& {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\

& \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x..........\left( 2 \right) \\

\end{align}$

Substituting ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ from equation $\left( 2 \right)$ in equation $\left( 1

\right)$, we get,

${{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30...........\left( 3 \right)$

We have a property of exponents,

${{a}^{b-c}}=\dfrac{{{a}^{b}}}{{{a}^{c}}}$, where $a,b,c\in R$.

Using this property in equation $\left( 3 \right)$ with $a=81,b=1$ and $c={{\sin }^{2}}x$, we get,

${{81}^{{{\sin }^{2}}x}}+\dfrac{{{81}^{1}}}{{{81}^{{{\sin }^{2}}x}}}=30...........\left( 4 \right)$

Now, let us assume ${{81}^{{{\sin }^{2}}x}}=t$. Substituting ${{81}^{{{\sin }^{2}}x}}=t$ in equation

$\left( 4 \right)$, we get,

$t+\dfrac{81}{t}=30$

$\begin{align}

& \Rightarrow \dfrac{{{t}^{2}}+81}{t}=30 \\

& \Rightarrow {{t}^{2}}+81=30t \\

& \Rightarrow {{t}^{2}}-30t+81=0 \\

\end{align}$

Now, we have got a quadratic equation which can be easily solved by using the method of splitting the middle term. Solving this quadratic equation,

\[\begin{align}

& {{t}^{2}}-3t-27t+81=0 \\

& \Rightarrow t\left( t-3 \right)-27\left( t-3 \right)=0 \\

& \Rightarrow \left( t-3 \right)\left( t-27 \right)=0 \\

\end{align}\]

$\Rightarrow t=27$ or $t=3.........\left( 5 \right)$

Since we have earlier assumed ${{81}^{{{\sin }^{2}}x}}=t$, we can now again substitute

$t={{81}^{{{\sin }^{2}}x}}$ in equation $\left( 5 \right)$. Substituting $t={{81}^{{{\sin }^{2}}x}}$ in

equation $\left( 5 \right)$, we get,

${{81}^{{{\sin }^{2}}x}}=27$ or ${{81}^{{{\sin }^{2}}x}}=3$

Since, the above two equation contains exponential terms, in order to solve them, we first have to

make their base equal. We know $81={{3}^{4}}$ and $27={{3}^{3}}$. Hence, substituting

$81={{3}^{4}}$ and $27={{3}^{3}}$ in the above equation, we get,

${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}={{3}^{3}}$ or ${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}=3$

$\Rightarrow {{3}^{4{{\sin }^{2}}x}}={{3}^{3}}$ or ${{3}^{4}}^{{{\sin }^{2}}x}=3............\left( 6 \right)$

Now, the base of all the exponential terms in the above two equations in equation $\left( 6 \right)$ is the same. So, we can directly equate the powers of the terms on both the sides of equality.

$\Rightarrow 4{{\sin }^{2}}x=3$ or $4{{\sin }^{2}}x=1$

$\Rightarrow {{\sin }^{2}}x=\dfrac{3}{4}$ or ${{\sin }^{2}}x=\dfrac{1}{4}$

$\Rightarrow \sin x=\pm \dfrac{\sqrt{3}}{2}$ or $\sin x=\pm \dfrac{1}{2}.........\left( 7 \right)$

In the question, it is given that \[0\le x\le \pi \]. For this domain, the value of $\sin x$ is positive .

So, we can neglect $\sin x=-\dfrac{\sqrt{3}}{2}$ and $\sin x=-\dfrac{1}{2}$

as the solution of the equation $\left( 7 \right)$.

Removing $\sin x=-\dfrac{\sqrt{3}}{2}$ and $\sin x=-\dfrac{1}{2}$ from equation $\left( 7 \right)$, we

get,

$\sin x=+\dfrac{\sqrt{3}}{2}$ or $\sin x=+\dfrac{1}{2}$

Solving the above equation in the domain \[0\le x\le \pi \], we get,

$x=\dfrac{\pi }{3},\dfrac{2\pi }{3}$ or $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$.

So, the possible values of $x$ are $\dfrac{\pi }{3},\dfrac{\pi }{6},\dfrac{2\pi }{3},\dfrac{5\pi }{6}$.

Therefore the final answer is option (e).

Note: There is a possibility that one may commit a mistake if he/she does not consider $\dfrac{2\pi}{3}$ and $\dfrac{5\pi }{6}$ as possible solutions of the equation. But as it is mentioned in the question that \[0\le x\le \pi \], we have to consider $\dfrac{2\pi }{3}$ and $\dfrac{5\pi }{6}$ as possible solutions of the equation.

In the question, we are given an equation ${{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos}^{2}}x}}=30.........(1)$.

We have a trigonometric identity,

$\begin{align}

& {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\

& \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x..........\left( 2 \right) \\

\end{align}$

Substituting ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ from equation $\left( 2 \right)$ in equation $\left( 1

\right)$, we get,

${{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30...........\left( 3 \right)$

We have a property of exponents,

${{a}^{b-c}}=\dfrac{{{a}^{b}}}{{{a}^{c}}}$, where $a,b,c\in R$.

Using this property in equation $\left( 3 \right)$ with $a=81,b=1$ and $c={{\sin }^{2}}x$, we get,

${{81}^{{{\sin }^{2}}x}}+\dfrac{{{81}^{1}}}{{{81}^{{{\sin }^{2}}x}}}=30...........\left( 4 \right)$

Now, let us assume ${{81}^{{{\sin }^{2}}x}}=t$. Substituting ${{81}^{{{\sin }^{2}}x}}=t$ in equation

$\left( 4 \right)$, we get,

$t+\dfrac{81}{t}=30$

$\begin{align}

& \Rightarrow \dfrac{{{t}^{2}}+81}{t}=30 \\

& \Rightarrow {{t}^{2}}+81=30t \\

& \Rightarrow {{t}^{2}}-30t+81=0 \\

\end{align}$

Now, we have got a quadratic equation which can be easily solved by using the method of splitting the middle term. Solving this quadratic equation,

\[\begin{align}

& {{t}^{2}}-3t-27t+81=0 \\

& \Rightarrow t\left( t-3 \right)-27\left( t-3 \right)=0 \\

& \Rightarrow \left( t-3 \right)\left( t-27 \right)=0 \\

\end{align}\]

$\Rightarrow t=27$ or $t=3.........\left( 5 \right)$

Since we have earlier assumed ${{81}^{{{\sin }^{2}}x}}=t$, we can now again substitute

$t={{81}^{{{\sin }^{2}}x}}$ in equation $\left( 5 \right)$. Substituting $t={{81}^{{{\sin }^{2}}x}}$ in

equation $\left( 5 \right)$, we get,

${{81}^{{{\sin }^{2}}x}}=27$ or ${{81}^{{{\sin }^{2}}x}}=3$

Since, the above two equation contains exponential terms, in order to solve them, we first have to

make their base equal. We know $81={{3}^{4}}$ and $27={{3}^{3}}$. Hence, substituting

$81={{3}^{4}}$ and $27={{3}^{3}}$ in the above equation, we get,

${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}={{3}^{3}}$ or ${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}=3$

$\Rightarrow {{3}^{4{{\sin }^{2}}x}}={{3}^{3}}$ or ${{3}^{4}}^{{{\sin }^{2}}x}=3............\left( 6 \right)$

Now, the base of all the exponential terms in the above two equations in equation $\left( 6 \right)$ is the same. So, we can directly equate the powers of the terms on both the sides of equality.

$\Rightarrow 4{{\sin }^{2}}x=3$ or $4{{\sin }^{2}}x=1$

$\Rightarrow {{\sin }^{2}}x=\dfrac{3}{4}$ or ${{\sin }^{2}}x=\dfrac{1}{4}$

$\Rightarrow \sin x=\pm \dfrac{\sqrt{3}}{2}$ or $\sin x=\pm \dfrac{1}{2}.........\left( 7 \right)$

In the question, it is given that \[0\le x\le \pi \]. For this domain, the value of $\sin x$ is positive .

So, we can neglect $\sin x=-\dfrac{\sqrt{3}}{2}$ and $\sin x=-\dfrac{1}{2}$

as the solution of the equation $\left( 7 \right)$.

Removing $\sin x=-\dfrac{\sqrt{3}}{2}$ and $\sin x=-\dfrac{1}{2}$ from equation $\left( 7 \right)$, we

get,

$\sin x=+\dfrac{\sqrt{3}}{2}$ or $\sin x=+\dfrac{1}{2}$

Solving the above equation in the domain \[0\le x\le \pi \], we get,

$x=\dfrac{\pi }{3},\dfrac{2\pi }{3}$ or $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$.

So, the possible values of $x$ are $\dfrac{\pi }{3},\dfrac{\pi }{6},\dfrac{2\pi }{3},\dfrac{5\pi }{6}$.

Therefore the final answer is option (e).

Note: There is a possibility that one may commit a mistake if he/she does not consider $\dfrac{2\pi}{3}$ and $\dfrac{5\pi }{6}$ as possible solutions of the equation. But as it is mentioned in the question that \[0\le x\le \pi \], we have to consider $\dfrac{2\pi }{3}$ and $\dfrac{5\pi }{6}$ as possible solutions of the equation.

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