# 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

Answer

Verified

364.2k+ views

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.

Last updated date: 27th Sep 2023

â€¢

Total views: 364.2k

â€¢

Views today: 5.64k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the basic unit of classification class 11 biology CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Draw a welllabelled diagram of a plant cell class 11 biology CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE