Question

How do you solve $4{{\sin }^{2}}x=1$ for $x$ in the interval $\left[ 0,2\pi \right)$?

Answer 1

Hint: We have been given a quadratic equation of $\sin x$. We divide both sides of the equation by the constant 4. Then we take the square root on both sides of the equation. From that we find the exact solutions for the equation $4{{\sin }^{2}}x=1$ for $x$ in the interval $\left[ 0,2\pi \right)$.

Complete step by step answer:
The given equation of $\sin x$ is $4{{\sin }^{2}}x=1$.
We divide both sides of the equation by the constant 4 and get
$\begin{align}
  & \dfrac{4{{\sin }^{2}}x}{4}=\dfrac{1}{4} \\
 & \Rightarrow {{\sin }^{2}}x=\dfrac{1}{4} \\
\end{align}$
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\sin }^{2}}x}=\sqrt{\dfrac{1}{4}} \\
 & \Rightarrow \sin x=\pm \dfrac{1}{2} \\
\end{align}$
We know that in the principal domain or the periodic value of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ for $\sin x$, if we get $\sin a=\sin b$ where $-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2}$ then $a=b$.
We have $\sin x=\dfrac{1}{2}$, the value of \[\sin \left( \dfrac{\pi }{6} \right),\sin \left( \dfrac{5\pi }{6} \right)\] as $\dfrac{1}{2}$ in the domain of $\left[ 0,2\pi \right)$.
We have $\sin x=-\dfrac{1}{2}$, the value of \[\sin \left( \dfrac{7\pi }{6} \right),\sin \left( \dfrac{11\pi }{6} \right)\] as $-\dfrac{1}{2}$ in the domain of $\left[ 0,2\pi \right)$.
Therefore, $\sin x=\pm \dfrac{1}{2}$ gives $x=\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{7\pi }{6},\dfrac{11\pi }{6}$ as primary value.

Note:
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$. For our given problem $\sin x=\pm \dfrac{1}{2}$, the primary solution is $x=\pm \dfrac{\pi }{6}$.
The general solution will be $x=\left( n\pi \pm {{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right)$. Here $n\in \mathbb{Z}$.