Question

How do you solve $\csc x + \cot x = 1$ and find all solutions in the interval $\left[ {0,2\pi } \right)$?

Answer 1

Here, we will first convert the given trigonometric function into sine and cosine functions. We will then use basic mathematical formulas and suitable algebraic identity to convert the equation into a quadratic equation. Then we will solve the quadratic equation to find the required solution of the given equation.

Formula Used:
We will use the following formula:
1. Trigonometric Ratio: $\cot x = \dfrac{{\cos x}}{{\sin x}}$
2. Trigonometric Co-ratio: $\csc x = \dfrac{1}{{\sin x}}$
3. Trigonometric identity: ${\sin ^2}x + {\cos ^2}x = 1$

Complete Step by Step Solution:
We are given an equation $\csc x + \cot x = 1$.
We know that the Trigonometric Ratio: $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and Trigonometric Co-ratio: $\csc x = \dfrac{1}{{\sin x}}$
Now, by rewriting the equation in terms of sine and cosine using the Trigonometric ratio and Trigonometric Co-ratio, we get
$\Rightarrow \dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} = 1$
By adding the numerators in the fraction, we get
$\Rightarrow \dfrac{{1 + \cos x}}{{\sin x}} = 1$
By rewriting the equation, we get
$\Rightarrow 1 + \cos x = \sin x$
By squaring on both the sides of the equation, we get
$\Rightarrow {\left( {1 + \cos x} \right)^2} = {\left( {\sin x} \right)^2}$
The square of the sum of the numbers is given by an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Now, by using the algebraic identity, we get
$\Rightarrow 1 + {\left( {\cos x} \right)^2} + 2\cos x = {\left( {\sin x} \right)^2}$
$\Rightarrow 1 + {\cos ^2}x + 2\cos x = {\sin ^2}x$
We know that Trigonometric identity: ${\sin ^2}x + {\cos ^2}x = 1$
$\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Now, by using the Trigonometric identity, we get
$1 + {\cos ^2}x + 2\cos x = 1 - {\cos ^2}x$
By rewriting the equation, we get
$\Rightarrow 1 + {\cos ^2}x + 2\cos x - 1 + {\cos ^2}x = 0$
Adding and subtracting the like terms, we get
$\Rightarrow 2{\cos ^2}x + 2\cos x = 0$
By taking out the common factors, we get
$\Rightarrow 2\cos x\left( {\cos x + 1} \right) = 0$
By using the zero product property, we get
$\Rightarrow 2\cos x = 0$ or $\cos x + 1 = 0$
$\Rightarrow \cos x = 0$ or $\cos x = - 1$
$\Rightarrow x = {\cos ^{ - 1}}\left( 0 \right)$ or $x = {\cos ^{ - 1}}\left( { - 1} \right)$
We know that $\cos n\dfrac{\pi }{2} = 0$ where$n$ is odd and $\cos \pi = - 1$ , we get
$\Rightarrow x = n\dfrac{\pi }{2}$ or $x = \pi$
Since it is given that the solutions of the equations should lie in the interval $\left[ {0,\left. {2\pi } \right)} \right.$, so we get
$\Rightarrow x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$ or $x = \pi$ where $n = 1,3$ and $x \in \left[ {0,2\pi } \right)$
When $x = \dfrac{{3\pi }}{2}$ and $x = \pi$, then the Trigonometric equation $\csc x + \cot x = 1$is not defined.
The only solution satisfying the given Trigonometric equation $\csc x + \cot x = 1$ is $x = \dfrac{\pi }{2}$ where $x \in \left[ {0,2\pi } \right)$.

Therefore, the solutions of the equation $\csc x + \cot x = 1$ in the interval $\left[ {0,\left. {2\pi } \right)} \right.$ is $\dfrac{\pi }{2}$.

Note:
We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation that is always true for all the variables. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right-angled triangle with respect to any of its acute angle. We know that the inverse trigonometric function is used to find the missing angles in a right-angled triangle. We should know that an Open interval is an interval that does not include the endpoints and is denoted by () whereas a Closed interval is an interval that includes the endpoints and is denoted by [].Zero product property states that when the product of two factors is zero, then one of the factor is separately zero i.e., if $ab = 0$ then $a = 0$ or $b = 0$.