Question

How do you solve: $\cos ecx + \cot x = 1$ for $0 \leqslant x \leqslant 2\pi $ ?

Answer 1

Hint:The given question involves solving a trigonometric equation and finding the value of angle x that satisfies the given trigonometric equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step by step answer:
The given problem requires us to solve the trigonometric equation $\cos ecx + \cot x = 1$.The given trigonometric equation can be solved by first converting the given trigonometric ratios into cosine and sine and then condensing both the terms into a single compound angle sine or cosine formula. So, we know that $\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$. Hence, we get,
$\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} = 1$
$ \Rightarrow \dfrac{{1 + \cos x}}{{\sin x}} = 1$
Since sinx is present in the denominator, so sinx cannot be equal to zero.
So, $\sin x \ne 0$. Now, simplifying the original expression further.
Shifting the sine from left side of the equation to right side of the equation, we get,
$ \Rightarrow 1 + \cos x = \sin x$
Shifting the cosine from left side of the equation to right side of the equation, we get,
$ \Rightarrow \sin x - \cos x = 1$
Dividing both sides of the equation by $\sqrt 2 $ to form a condensed trigonometric formula, we get,
$ \Rightarrow \dfrac{{\sin \left( x \right)}}{{\sqrt 2 }} - \dfrac{{\cos \left( x \right)}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}$
We know that $\sin \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$, we get,
$ \Rightarrow \sin \left( x \right)\cos \left( {\dfrac{\pi }{4}} \right) - \cos \left( x \right)\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
Using the compound formulae of sine and cosine, $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ and $\cos (A - B) = \cos A\cos B + \sin A\sin B$, we get,
$ \Rightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)$
Now, we know that general solution for the equation $\sin \left( \theta \right) = \sin \left( \phi \right)$ is $\theta = n\pi + {\left( { - 1} \right)^n}\phi $, where n is any integer.
$ \Rightarrow \left( {x - \dfrac{\pi }{4}} \right) = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right)$, where n is an integer.
Taking $\left( {\dfrac{\pi }{4}} \right)$ to right side of the equation, we get,
\[ x = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}\] , where n is any integer.
So, solution of the given equation $\cos ecx + \cot x = 1$ is: $x = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}$, where n is any integer.
So, putting $n = 0$, we get,
$ \Rightarrow x = \left( 0 \right)\pi + {\left( { - 1} \right)^0}\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}$
$ \Rightarrow x = \left( {\dfrac{\pi }{2}} \right)$
So, putting $n = 1$, we get,
$ \Rightarrow x = \left( 1 \right)\pi + {\left( { - 1} \right)^1}\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}$
$ \Rightarrow x = \pi + \left( { - 1} \right)\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}$
$ \therefore x = \pi $
So, we get values of x satisfying the given trigonometric equation as: $x = \pi $and $x = \dfrac{\pi }{2}$. But $\sin x$ should not be equal to zero for x to satisfy the original trigonometric equation given to us. Hence, the value of x cannot be zero.

Hence, the only possible value of x is $\dfrac{\pi }{2}$.

Note:Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. The general solution of a given trigonometric solution may differ in form, but actually represents the correct solutions. The different forms of general equations are interconvertible into each other. For solving such types of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.