Question

The equation of $2\cot 2x-3\cot 3x=\tan 2x$ has
A. Two solutions in $\left( 0,\dfrac{\pi }{3} \right)$.
B. One solution in $\left( 0,\dfrac{\pi }{3} \right)$.
C. No solution in $\left( -\infty ,\infty \right)$.
D. None of these

Answer 1

Hint: In this problem we have found the solution for the given equation that is $2\cot 2x-3\cot 3x=\tan 2x$. In this problem we observe the given equation has trigonometric ratios. In the question we have an expression in $\cot $. So we will change $\cot $ into $\tan $ by using the well-known trigonometric formula $\cot x=\dfrac{1}{\tan x}$. Now we will simplify the equation by taking LCM. Then we will get a quadratic equation. To solve the quadratic equation, we will use the quadratic equation formula and simplify it to get the solution for the given equation.

Formula used:
1.$\cot x=\dfrac{1}{\tan x}$.
2.$\tan 2x=\dfrac{2\tan x}{\left( 1-{{\tan }^{2}}x \right)}$.
3. $\tan 3x=\dfrac{3\tan x-{{\tan }^{3}}x}{\left( 1-3{{\tan }^{2}}x \right)}$.
4. Solution for the quadratic equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{b-4ac}}{2ac}$.

Complete Step by Step Solution:
Given that, $2\cot 2x-3\cot 3x=\tan 2x$.
Now we will change the $\cot x$ into $\tan x$by using the formula $\cot x=\dfrac{1}{\tan x}$, then we will write
$\Rightarrow \dfrac{2}{\tan 2x}-\dfrac{3}{\tan 3x}=\tan 2x$.
We know that trigonometric identity that is $\tan 2x=\dfrac{2{{\tan }^{2}}x}{1-{{\tan }^{2}}x}$. Substituting this formula in the above equation, then we will get
$\Rightarrow \dfrac{2\left( 1-{{\tan }^{2}}x \right)}{2\tan x}-\dfrac{3}{\tan 3x}=\tan 2x$.
We have another trigonometric identity that is $\tan 3x=\dfrac{3\tan x-{{\tan }^{3}}x}{\left( 1-3{{\tan }^{2}}x \right)}$ . Again, substituting this formula in the above equation, then we will get
 $\Rightarrow \dfrac{2\left( 1-{{\tan }^{2}}x \right)}{2\tan x}-\dfrac{3\left( 1-3{{\tan }^{2}}x \right)}{3\tan x-{{\tan }^{3}}x}=\tan 2x$.
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow \dfrac{2\left( 1-{{\tan }^{2}}x \right)\left( 3-\tan 2x+{{\tan }^{4}}x-3+9{{\tan }^{2}}x \right)}{\tan x\left( 3-{{\tan }^{2}}x \right)}=\dfrac{2\tan x}{1-{{\tan }^{2}}x} \\
 & \Rightarrow \dfrac{{{\tan }^{{}}}x+5{{\tan }^{2}}x}{\tan x\left( 3-{{\tan }^{2}}x \right)}=\dfrac{2\tan x}{1-{{\tan }^{2}}x} \\
 & \Rightarrow \left( {{\tan }^{2}}x+5 \right)\left( 1-{{\tan }^{2}}x \right)=2\left( 3-{{\tan }^{2}}x \right) \\
 & \Rightarrow {{\tan }^{4}}x+4{{\tan }^{2}}x+1=0 \\
\end{align}$
We can observe that the above equation is quadratic equation in terms of ${{\tan }^{2}}x$, so the solution of the above equation can be written as
$\begin{align}
  & {{\tan }^{2}}x=\dfrac{-2\pm \sqrt{4-4}}{2} \\
 & \Rightarrow {{\tan }^{2}}x=-1 \\
\end{align}$
 we will get the negative solution. Then there is no solution.

Note:
We can also convert the whole given equation in terms of $\sin x$, $\cos x$ by using the well-known formulas $\tan x=\dfrac{\sin x}{\cos x}$, $\cot x=\dfrac{\cos x}{\sin x}$. After substituting these formulas, we will simplify the equation and use trigonometric identities to solve the equation.