Question

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 1

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,
$\begin{aligned} & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x..........\left( 2 \right) \\\ \end{aligned}$
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{aligned}

& \Rightarrow \dfrac{{{t}^{2}}+81}{t}=30 \\
& \Rightarrow {{t}^{2}}+81=30t \\
& \Rightarrow {{t}^{2}}-30t+81=0 \\
\end{aligned}$
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,

& {{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{aligned}$$ $\Rightarrow t=27$ or $t=3.........\left( 5 \right)$ Since we have earlier assumed ${{81}^{{{\sin }^{2}}x}}=t$, we can now again substitute $t={{81}^{{{\sin }^{2}}x}}$ in equation $\left( 5 \right)$. Substituting $t={{81}^{{{\sin }^{2}}x}}$ in equation $\left( 5 \right)$, we get, ${{81}^{{{\sin }^{2}}x}}=27$ or ${{81}^{{{\sin }^{2}}x}}=3$ Since, the above two equation contains exponential terms, in order to solve them, we first have to make their base equal. We know $81={{3}^{4}}$ and $27={{3}^{3}}$. Hence, substituting $81={{3}^{4}}$ and $27={{3}^{3}}$ in the above equation, we get, ${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}={{3}^{3}}$ or ${{\left( {{3}^{4}} \right)}^{{{\sin }^{2}}x}}=3$ $\Rightarrow {{3}^{4{{\sin }^{2}}x}}={{3}^{3}}$ or ${{3}^{4}}^{{{\sin }^{2}}x}=3............\left( 6 \right)$ Now, the base of all the exponential terms in the above two equations in equation $\left( 6 \right)$ is the same. So, we can directly equate the powers of the terms on both the sides of equality. $\Rightarrow 4{{\sin }^{2}}x=3$ or $4{{\sin }^{2}}x=1$ $\Rightarrow {{\sin }^{2}}x=\dfrac{3}{4}$ or ${{\sin }^{2}}x=\dfrac{1}{4}$ $\Rightarrow \sin x=\pm \dfrac{\sqrt{3}}{2}$ or $\sin x=\pm \dfrac{1}{2}.........\left( 7 \right)$ In the question, it is given that $$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.