Question

How do I solve $4\sin x + 9\cos x = 0$ for $0 < x < {360^ \circ }$?

Answer 1

Hint: First, subtract $9\cos x$ from both sides of the equation. Then, divide both sides of the equation by $4\cos x$ and simplify using trigonometry identities. Then, find the value of $x$ satisfying $\tan x = - \dfrac{9}{4}$ using trigonometric properties. Next, find all values of $x$ in the interval $0 < x < {360^ \circ }$. Then, we will get all the solutions of the given equation in the given interval.
Formula used:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
${\tan ^{ - 1}}\left( {\dfrac{9}{4}} \right) = 1.15$
$\tan \left( {\pi - x} \right) = - \tan x$
$\tan \left( {2\pi - x} \right) = - \tan x$

Complete step by step answer:
Given equation: $4\sin x + 9\cos x = 0$
We have to find all possible values of $x$ satisfying a given equation in the interval $0 < x < {360^ \circ }$.
First, subtract $9\cos x$ from both sides of the equation.
$4\sin x = - 9\cos x$
Divide both sides of the equation by $4\cos x$, we get
$\dfrac{{\sin x}}{{\cos x}} = - \dfrac{9}{4}$
Now, use the identity $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ in the above equation.
$ \Rightarrow \tan x = - \dfrac{9}{4}$
Now, we will find the values of $x$ satisfying $\tan x = - \dfrac{9}{4}$…(i)
So, using the property $\tan \left( {\pi - x} \right) = - \tan x$ and ${\tan ^{ - 1}}\left( {\dfrac{9}{4}} \right) = 1.15$ in equation (i).
$ \Rightarrow \tan \left( x \right) = - \tan \left( {1.15} \right)$
$ \Rightarrow \tan \left( x \right) = \tan \left( {3.14 - 1.15} \right)$
$ \Rightarrow x = 1.99$
Now, using the property $\tan \left( {2\pi - x} \right) = - \tan x$ and ${\tan ^{ - 1}}\left( {\dfrac{9}{4}} \right) = 1.15$ in equation (i).
$ \Rightarrow \tan \left( x \right) = - \tan \left( {1.15} \right)$
$ \Rightarrow \tan \left( x \right) = \tan \left( {2 \times 3.14 - 1.15} \right)$
$ \Rightarrow x = 5.13$
Since, the period of the $\tan \left( x \right)$ function is $\pi $ so values will repeat every $\pi $ radians in both directions.
$x = 1.99 + n\pi ,5.13 + n\pi $, for any integer $n$.
Now, find all values of $x$ in the interval $0 < x < {360^ \circ }$.
Since, it is given that $x \in \left( {0,6.28} \right)$, hence put $n = 0$ in the general solution.
So, putting $n = 0$ in the general solution, $x = 1.99 + n\pi ,5.13 + n\pi $, we get
$\therefore x = 1.99,5.13$
Final solution: Hence, $x = 1.99,5.13$ are the solutions of the given equation in the given interval.

Note:
In above question, we can find the solutions of given equation by plotting the equation, $4\sin x + 9\cos x = 0$ on graph paper and determine all solutions which lie in the interval, $0 < x < {360^ \circ }$.

From the graph paper, we can see that there are two values of $x$ in the interval $0 < x < {360^ \circ }$.
So, these will be the solutions of the given equation in the given interval.
Final solution: Hence, $x = 1.99,5.13$ are the solutions of the given equation in the given interval.