Question

Solve ${\csc ^2}x - 2{\cot ^2}x = 0$

Answer 1

Hint: Now in order to solve the above equation we should work with one side at a time and manipulate it to the other side. Using some basic trigonometric identities like we can simplify the above expression.
$
\csc x = \dfrac{1}{{\sin x}} \\
\cot x = \dfrac{{\cos x}}{{\sin x}} \\
$
Such that in order to solve the given expression we have to use the above identities and express our given expression in that form and thereby verify it.

Complete step by step solution:
Given
${\csc ^2}x - 2{\cot ^2}x = 0...........................\left( i \right)$
Here we have to basically solve for $x$ using any mathematical operations and functions. In order to solve for $x$ we can use the above mentioned trigonometric identities to simplify it.
Now we know that:
$
\csc x = \dfrac{1}{{\sin x}} \\
\cot x = \dfrac{{\cos x}}{{\sin x}} \\
$
So substituting the above expression in the given equation, we can write:
$
\Rightarrow {\csc ^2}x - 2{\cot ^2}x = 0 \\
\Rightarrow {\left( {\dfrac{1}{{\sin x}}} \right)^2} - 2{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} =
0.....................\left( {ii} \right) \\
$
Now we have to simplify the equation (ii). Such that both the terms have the same denominator. So we can write:
$
\Rightarrow {\left( {\dfrac{1}{{\sin x}}} \right)^2} - 2{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} = 0
\\
\Rightarrow \dfrac{{1 - 2{{\cos }^2}x}}{{{{\sin }^2}x}} = 0 \\
$
Now on substituting $\csc x = \dfrac{1}{{\sin x}}$ in the above equation we can write:
$
\Rightarrow \dfrac{{1 - 2{{\cos }^2}x}}{{{{\sin }^2}x}} = 0 \\
\Rightarrow {\csc ^2}x\left( {1 - 2{{\cos }^2}x} \right) = 0.......................\left( {iii} \right) \\
$
Now in order to solve for $x$ we have to equate each term in the LHS to the RHS which is 0 and thereby find the possible values of $x$.
So we can write:
\[
{\csc ^2}x\left( {1 - 2{{\cos }^2}x} \right) = 0 \\
\Rightarrow {\csc ^2}x = 0\;\;\;\;{\text{and}}\;\;\; \Rightarrow 1 - 2{\cos ^2}x = 0 \\
\Rightarrow {\csc ^2}x = 0\;\;\;\;{\text{and}}\;\;\; \Rightarrow {\cos ^2}x = \dfrac{1}{2} \\
\Rightarrow {\csc ^2}x = 0\;\;\;\;{\text{and}}\;\;\; \Rightarrow \cos x = \pm \sqrt {\dfrac{1}{2}}
\\
\Rightarrow x = {\text{not defined}}\;\;\;\;{\text{and}}\;\;\; \Rightarrow \cos x = \pm \sqrt
{\dfrac{1}{2}} \\
\Rightarrow x = {\text{not defined}}\;\;\;\;{\text{and}}\;\;\; \Rightarrow x = n\pi + \dfrac{\pi
}{4}\;\;\;n = 0,1,2......... \\
\]
Therefore on solving ${\csc ^2}x - 2{\cot ^2}x = 0$ we get\[x = n\pi + \dfrac{\pi }{4}\;\].

Note: Some other equations needed for solving these types of problem are:
\[
1 + {\tan ^2}x = {\sec ^2}x \\
\begin{array}{*{20}{l}}
{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\
{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin}^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1}
\end{array} \\
\]
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.