Question

Find the possible value of $\cos x$, if $\cot x + \cos ecx = 5$.

Answer 1

Here, we are required to find the possible value of $\cos x$. We will write $\cot x$ and $\cos ecx$ in terms of $\sin x$ and $\cos x$. Then using various properties we will simplify the equation. We will then obtain the equation in quadratic form. We will simplify the equation using the factorization method and find the required answer.

Formulas Used:
We will use the following formulas:

1. ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
2. $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
3. $\left( {\cos e{c^2}x - {{\cot }^2}x} \right) = 1$
4. ${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step solution:
According to the question,
$\cot x + \cos ecx = 5$…………………………$\left( 1 \right)$
As we know, $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$
Therefore, substituting $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$ in equation (1), we get
$\Rightarrow \dfrac{{\cos x}}{{\sin x}} + \dfrac{1}{{\sin x}} = 5$
Since the denominator in the LHS is same, therefore, adding the numerators, we get
$\dfrac{{\cos x + 1}}{{\sin x}} = 5$……………………………..$\left( 2 \right)$
On cross multiplication, we get
$\Rightarrow \cos x + 1 = 5\sin x$
Now, squaring both sides, we get
$\Rightarrow {\left( {\cos x + 1} \right)^2} = {\left( {5\sin x} \right)^2}$
We will use the formula${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to simplify LHS of the above equation. Therefore, we get
$\Rightarrow {\cos ^2}x + 2\cos x + 1 = 25{\sin ^2}x$…………………………………..$\left( 3 \right)$
Now, we know that ${\sin ^2}x + {\cos ^2}x = 1$, hence,
${\sin ^2}x = 1 - {\cos ^2}x$
Substituting ${\sin ^2}x = 1 - {\cos ^2}x$ in equation$\left( 3 \right)$ we get,
$\Rightarrow {\cos ^2}x + 2\cos x + 1 = 25\left( {1 - {{\cos }^2}x} \right)$
$\Rightarrow {\cos ^2}x + 2\cos x + 1 = 25 - 25{\cos ^2}x$
Subtracting $25 - 25{\cos ^2}x$ from both sides, we get
Rearranging the terms, we get
$\Rightarrow 26{\cos ^2}x + 2\cos x - 24 = 0$
Dividing by 2 on both the sides, we get
$\Rightarrow 13{\cos ^2}x + \cos x - 12 = 0$
The above equation is in the form of a quadratic equation.
Factoring the equation, we get
$\Rightarrow 13{\cos ^2}x + 13\cos x - 12\cos x - 12 = 0$
Factoring out common term, we get
$\Rightarrow 13\cos x\left( {\cos x + 1} \right) - 12\left( {\cos x + 1} \right) = 0$
$\Rightarrow \left( {\cos x + 1} \right)\left( {13\cos x - 12} \right) = 0$
Using zero product property, we get
$\begin{array}{l} \Rightarrow \cos x + 1 = 0\\\ \Rightarrow \cos x = - 1\end{array}$
Or
$\begin{array}{l} \Rightarrow \left( {13\cos x - 12} \right) = 0\\\ \Rightarrow \cos x = \dfrac{{12}}{{13}}\end{array}$
Now, we will reject the value $\cos x = - 1$ because from equation $\left( 2 \right)$,
$\dfrac{{\cos x + 1}}{{\sin x}} = 5$
Hence, if $\cos x = - 1$ then, LHS will become 0 which will not be equal to RHS.
Therefore, we will reject this value.

Hence, if $\cot x + \cos ecx = 5$, then the possible value of $\cos x$ is $\dfrac{{12}}{{13}}$.

Note:
This question can be solved in another way using the formula:
$\left( {\cos e{c^2}x - {{\cot }^2}x} \right) = 1$……………………………………..$\left( 5 \right)$
Now, it is given that $\cot x + \cos ecx = 5$………………..$\left( 1 \right)$
Using the formula $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ in the equation $\left( 5 \right)$, we get
$\Rightarrow \left( {\cos ecx - \cot x} \right)\left( {\cos ecx + \cot x} \right) = 1$
Substituting $\cot x + \cos ecx = 5$ in the above equation, we get
$\Rightarrow \left( {\cos ecx - \cot x} \right)\left( 5 \right) = 1$
$\Rightarrow \left( {\cos ecx - \cot x} \right) = \dfrac{1}{5}$……………………….$\left( 6 \right)$
Adding $\left( 1 \right)$ and $\left( 6 \right)$, we get
$\cot x + \cos ecx + \cos ecx - \cot x = \dfrac{1}{5} + 5 = \dfrac{{26}}{5}$
$\Rightarrow 2\cos ecx = \dfrac{{26}}{5}$
Dividing both sides by 2, we get
$\Rightarrow \cos ecx = \dfrac{{13}}{5}$
Now we know $\sin x = \dfrac{1}{{\cos ecx}}$, so
$\Rightarrow \sin x = \dfrac{5}{{13}}$
Now, we have shown above that ${\sin ^2}x = 1 - {\cos ^2}x$.
Taking square root on both sides of ${\sin ^2}x = 1 - {\cos ^2}x$, we get
$\sin x = \sqrt {1 - {{\cos }^2}x}$
Substituting $\sin x = \sqrt {1 - {{\cos }^2}x}$ in $\sin x = \dfrac{5}{{13}}$, we get
$\Rightarrow \dfrac{5}{{13}} = \sqrt {1 - {{\cos }^2}x}$
Squaring both sides, we get
$\Rightarrow \dfrac{{25}}{{169}} = 1 - {\cos ^2}x$
$\Rightarrow {\cos ^2}x = 1 - \dfrac{{25}}{{169}} = \dfrac{{169 - 25}}{{169}}$
Simplifying the expression, we get
$\Rightarrow {\cos ^2}x = \dfrac{{144}}{{169}}$
Since, we know that 144 is the square of 12.
$\Rightarrow {\cos ^2}x = {\left( {\dfrac{{12}}{{13}}} \right)^2}$
$\therefore \cos x=\left( \dfrac{12}{13} \right)$
Therefore, if $\cot x + \cos ecx = 5$, then the possible value of $\cos x$ is $\dfrac{{12}}{{13}}$.
Hence, this is the required answer.