Question

How do you solve $4{\sin ^2}x + 1 = - 4\sin x$?

Answer 1

Hint:
First, we have to make the right side of the equation to be $0$. Then, factor the left side of the equation. Next, substitute $u$ for all occurrences of $\sin x$and factor using the perfect square rule. Next, replace all occurrences of $u$ with $\sin x$ and replace the left side with the factored expression. Next, set $\left( {2\sin x + 1} \right)$ equal to $0$ and solve for $x$ using trigonometric properties. Then, we will get all solutions of the given equation.

Formula used:
Perfect square trinomial rule: 1) ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$
2) $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
3) $\sin \left( {\pi + x} \right) = - \sin x$
4) $\sin \left( {2\pi - x} \right) = - \sin x$

Complete step by step solution:
Given equation: $4{\sin ^2}x + 1 = - 4\sin x$
We have to find all possible values of $x$ satisfying a given equation.
First, we have to make the right side of the equation to be $0$.
So, adding $4\sin x$ to both sides of the given equation
$4{\sin ^2}x + 4\sin x + 1 = 0$
We have to factor the left side of the equation.
So, first put $u = \sin x$, i.e., substitute $u$ for all occurrences of $\sin x$.
$ \Rightarrow 4{u^2} + 4u + 1$
Now, we have to factor using the perfect square rule.
So, rewrite $4{u^2}$ as ${\left( {2u} \right)^2}$.
$ \Rightarrow {\left( {2u} \right)^2} + 4u + 1$
Now, rewrite $1$ as ${1^2}$.
$ \Rightarrow {\left( {2u} \right)^2} + 4u + {1^2}$
Now, we have to check the middle term by multiplying $2ab$ and compare this result with the middle term in the original expression.
$2ab = 2 \cdot \left( {2u} \right) \cdot 1$
Simplifying this, we get
$2ab = 4u$
Now, we have to factor using the perfect square trinomial rule ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$, where $a = 2u$ and $b = 1$.
$ \Rightarrow {\left( {2u + 1} \right)^2}$
Now, replace all occurrences of $u$ with $\sin x$.
$ \Rightarrow {\left( {2\sin x + 1} \right)^2}$
Now, replace the left side with the factored expression.
$ \Rightarrow {\left( {2\sin x + 1} \right)^2} = 0$
Now, set $\left( {2\sin x + 1} \right)$ equal to $0$ and solve for $\sin x$.
Now, set the factor equal to $0$.
$ \Rightarrow 2\sin x + 1 = 0$
Now, subtract $1$ to both sides of the equation.
$ \Rightarrow 2\sin x = - 1$
Now, divide each term by $2$ and simplify.
$ \Rightarrow \sin x = - \dfrac{1}{2}$…(i)
Now, using the property $\sin \left( {\pi + x} \right) = - \sin x$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ in equation (i).
$ \Rightarrow \sin x = - \sin \dfrac{\pi }{6}$
$ \Rightarrow \sin x = \sin \left( {\pi + \dfrac{\pi }{6}} \right)$
$ \Rightarrow x = \dfrac{{7\pi }}{6}$
Now, using the property $\sin \left( {2\pi - x} \right) = - \sin x$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ in equation (i).
$ \Rightarrow \sin x = - \sin \dfrac{\pi }{6}$
$ \Rightarrow \sin x = \sin \left( {2\pi - \dfrac{\pi }{6}} \right)$
$ \Rightarrow x = \dfrac{{11\pi }}{6}$
Since, the period of the $\sin x$ function is $2\pi $ so values will repeat every $2\pi $ radians in both directions.
$x = \dfrac{{7\pi }}{6} + 2n\pi ,\dfrac{{11\pi }}{6} + 2n\pi $, for any integer $n$.

Final solution: Hence, $x = \dfrac{{7\pi }}{6} + 2n\pi ,\dfrac{{11\pi }}{6} + 2n\pi $, for any integer $n$are solutions of the given equation.

Note:
In the above question, we can find the solutions of a given equation by plotting the equation, $4{\sin ^2}x + 1 = - 4\sin x$ on graph paper and determine all its solutions.

From the graph paper, we can see that $x = \dfrac{{7\pi }}{6}$ and $x = \dfrac{{11\pi }}{6}$ are solution of given equation, and solution repeat every $2\pi $ radians in both directions.
So, these will be the solutions of the given equation.
Final solution: Hence, $x = \dfrac{{7\pi }}{6} + 2n\pi ,\dfrac{{11\pi }}{6} + 2n\pi $, for any integer $n$ are solutions of the given equation.