Question

Find the solution set of the equation, ${{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 3x+\sin x \right)={{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 2x \right)$
(a) $x=-\dfrac{\pi }{3}$
(b) $x=-\dfrac{5\pi }{3}$
(c) $x=\dfrac{2\pi }{3}$
(d) $x=\dfrac{\pi }{3}$

Answer 1

Hint: In this problem, we are to find the solution set of the equation ${{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 3x+\sin x \right)={{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 2x \right)$. So, to start with, we will check if the logarithmic function is well defined or not by analyzing the given bases. Thus we will be able to find the range where x is taking place. Again, we will use trigonometric identities to simplify and get the values of x. Putting them back we will also get the solution which are satisfying the conditions and we will have to conclude it as the right solution.

Complete step-by-step solution:
To start with, we are to find the solution set of ${{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 3x+\sin x \right)={{\log }_{\dfrac{-{{x}^{2}}-6x}{10}}}\left( \sin 2x \right)$.
Now, to define log, the term $\dfrac{-{{x}^{2}}-6x}{10}$ should be greater than 0.
So,
$\dfrac{-{{x}^{2}}-6x}{10}>0$
Simplifying,
$\Rightarrow -{{x}^{2}}-6x>0$
Taking the negative sign common,
$\Rightarrow {{x}^{2}}+6x<0$
Hence,
$\Rightarrow x\left( x+6 \right)<0$
So, x has the value in the range, $\left( -6,0 \right)$
And, also the value of $\dfrac{-{{x}^{2}}-6x}{10}$must be not equal to 1.
So, we have,
$\dfrac{-{{x}^{2}}-6x}{10}\ne 1$
$\Rightarrow -{{x}^{2}}-6x\ne 10$
Changing sides,
$\Rightarrow {{x}^{2}}+6x+10\ne 0$
Now, if we try to find the determinant, we get,
$D={{\left( 6 \right)}^{2}}-4.1.10$ $=36-40=-4$ which is less than 0.
So, from this $x\in \mathbb{R}$ .
Then, x will have a value in the range, $\left( -6,0 \right)$
Again, let us consider, $\dfrac{-{{x}^{2}}-6x}{10}=a$
Then we get, ${{\log }_{a}}\left( \sin 3x+\sin x \right)={{\log }_{a}}\left( \sin 2x \right)$
Hence we can say, $\sin 3x+\sin x=\sin 2x$
Using trigonometric identities,
$\Rightarrow 2\sin \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)=\sin 2x$
Simplifying,
$\Rightarrow 2\sin 2x\cos x=\sin 2x$
$\Rightarrow \sin 2x\left( 2\cos x-1 \right)=0$
Then, we get,
$\sin 2x=0$ or $2\cos x-1=0$
But from the definition of logarithmic functions , we also have, $\sin 2x>0$ and also $\sin 3x+\sin x>0$ ,
Thus, $\sin 2x=0$ does not hold anymore.
Then, we have, $2\cos x-1=0$as our equation.
$2\cos x-1=0$gives us,
$\Rightarrow 2\cos x=1$
Simplifying,
$\Rightarrow \cos x=\dfrac{1}{2}$
Again, $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$
$\Rightarrow \cos x=\cos \dfrac{\pi }{3}$
From the generalized formula,
$\Rightarrow x=2n\pi \pm \dfrac{\pi }{3},n\in \mathbb{Z}$
Again, we are also having the condition that, x will have a value in the range, $\left( -6,0 \right)$
So, putting $n=0$ ,
$\Rightarrow x=\pm \dfrac{\pi }{3}$, but for the given range, we have, $x=-\dfrac{\pi }{3}$
Again, putting $n=-1$ ,
$\Rightarrow x=-2\pi \pm \dfrac{\pi }{3}$, but for the given range, we have, $x=-2\pi +\dfrac{\pi }{3}=-\dfrac{5\pi }{3}$
So, the values of x will be, $-\dfrac{\pi }{3},-\dfrac{5\pi }{3}$
If we put the value, $x=-\dfrac{\pi }{3}$, we are getting sin 2x as a negative number which is not agreeing with the given conditions.
Hence the solution is, (b) $x=-\dfrac{5\pi }{3}$

Note: This is a question where we have used both logarithmic and trigonometric identities to solve the problem. The sum rule of sin function has been used as well as the base of the log to be positive fact is also used. While solving these kinds of problems we are to take care of the given range and then also put back the given value to check if they are satisfying all the given conditions or not.