Question

How do you solve $\cos x=\sin 2x$, within the interval $\left[ 0,2\pi \right)$?

Answer 1

Hint: We have been given a quadratic equation of $\sin x$. We use the multiple angle formula of $\sin 2x=2\sin x\cos x$. We take common terms out to form the multiplied forms. Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number. We find individual terms which give 0. We solve them to get the general solution.

Complete step by step answer:
The given equation of $\sin x$ is $\cos x=\sin 2x$. We try to convert $\sin 2x$ using the multiple angle formula of $\sin 2x=2\sin x\cos x$.
The revised form of the equation is $\cos x=2\sin x\cos x$.
We take all the terms in one side and get $\cos x-2\sin x\cos x=0$
We try to take the common numbers out.
For $\cos x-2\sin x\cos x$, we take $\cos x$ and get $\cos x\left( 1-2\sin x \right)$.
The equation becomes $\cos x\left( 1-2\sin x \right)=0$.
Therefore, $\cos x\left( 1-2\sin x \right)=0$ has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
So, either $\cos x=0$ or $\left( 1-2\sin x \right)=0$. $\left( 1-2\sin x \right)=0$ gives $\sin x=\dfrac{1}{2}$.
We know that in the principal domain or the periodic value of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ for $\sin x$ and $\cos x$, if we get $\sin a=\sin b$ where $-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2}$ then $a=b$.
We have $\cos x=0$, the value of $\cos \left( \dfrac{\pi }{2} \right),\cos \left( \dfrac{3\pi }{2} \right)$ as 0 in the domain of $\left[ 0,2\pi \right)$.
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 $\left[ 0,2\pi \right)$.
Therefore, $\sin x=\dfrac{1}{2}$ gives $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$ as primary value and $\cos x=0$ gives $x=\dfrac{\pi }{2},\dfrac{3\pi }{2}$ as primary value.

Therefore, the primary solution for $\cos x=\sin 2x$ is $x=\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{\pi }{2},\dfrac{3\pi }{2}$ in the domain $\left[ 0,2\pi \right)$.

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 $\sin x=\dfrac{1}{2}$, the primary solution is $x=\dfrac{\pi }{6}$.
The general solution will be $x=\left( n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right)$. Here $n\in \mathbb{Z}$.
Similarly, we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$. For our given problem $\cos x=0$, the primary solution is $x=\dfrac{\pi }{2}$.
The general solution will be $x=n\pi \pm \dfrac{\pi }{2}$. Here $n\in \mathbb{Z}$.