Question

How do you solve $\sin 3x=\cos 3x$ ? $$$$

Answer 1

We recall trigonometric equation, principal solution and general solution. We use a complementary reduction formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta$ for $\theta =3x$ to convert the cosine into sine. We then use the general solution for $\sin \theta =\sin \alpha$ as $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$. We then solve for $x$. $$$$

Complete step by step answer:
We know that a trigonometric equation is an equation with trigonometric functions with unknown arguments as measure of angles. When we are asked to solve a trigonometric equation we have to find all possible measures of unknown angles.
We know that the first solution of the trigonometric equation within the interval $\left[ 0,2\pi \right]$ is called principal solution and using periodicity all possible solutions obtained with integer $n$ are called general solutions. The general solution of the trigonometric equation $\sin \theta =\sin \alpha$ with principal solution $\theta =\alpha$ are given with arbitrary integer $z$ as
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
We are given the following trigonometric equation in sine and cosine as
$\sin 3x=\cos 3x$
We convert the cosine into sine using complimentary reduction formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta$ for $\theta =3x$ in the above step to have
$\sin 3x=\sin \left( \dfrac{\pi }{2}-3x \right)$
We find the general solution of the above equation taking principal solution $\alpha =\dfrac{\pi }{2}-3x$ as
$3x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2}-3x \right)$
If $n$ is an even integer we get${{\left( -1 \right)}^{n}}=1$ and we have

& 3x=n\pi +1\left( \dfrac{\pi }{2}-3x \right) \\\ & \Rightarrow 6x=\dfrac{\pi }{2}+n\pi \\\ & \Rightarrow x=\dfrac{\pi }{12}+n\dfrac{\pi }{6} \\\ \end{aligned}$$ If $n$ is odd integer we get${{\left( -1 \right)}^{n}}=-1$ and we have $$\begin{aligned} & 3x=n\pi +\left( -1 \right)\left( \dfrac{\pi }{2}-3x \right) \\\ & \Rightarrow 3x=n\pi -\dfrac{\pi }{2}+3x \\\ & \Rightarrow n=\dfrac{1}{2} \\\ \end{aligned}$$ Since $n$ is an integer the above result is a contradiction so $n$ cannot be an odd integer. So for $n=2m$ for any integer $m$ we have the solution as $$\begin{aligned} & x=\dfrac{\pi }{12}+2m\times \dfrac{\pi }{6} \\\ & \Rightarrow x=\dfrac{\pi }{12}+m\dfrac{\pi }{3} \\\ \end{aligned}$$ **Note:** We can alternatively solve using the general solution of $\tan \theta =\tan \alpha $ which is given by $\theta =n\pi +\alpha $. Let us divide the given equation $\sin 3x=\cos 3x$ by $\cos 3x$ assuming $\cos 3x\ne 0$ to have; $$\begin{aligned} & \dfrac{\sin 3x}{\cos 3x}=1 \\\ & \Rightarrow \tan 3x=1 \\\ & \Rightarrow \tan 3x=\tan \left( \dfrac{\pi }{4} \right) \\\ \end{aligned}$$ So the principal solution is $\alpha =\dfrac{\pi }{4}$ and the general solution is $$\begin{aligned} & \Rightarrow 3x=n\pi +\dfrac{\pi }{4} \\\ & \Rightarrow x=\dfrac{n\pi }{3}+\dfrac{\pi }{12} \\\ & \Rightarrow x=\dfrac{\pi }{12}+n\dfrac{\pi }{3} \\\ \end{aligned}$$ Since we have assumed $\cos 3x\ne 0$ we have $3x\ne \left( 2m+1 \right)\dfrac{\pi }{2}\Rightarrow x\ne \left( 2m+1 \right)\dfrac{\pi }{6}$ for any integer $m$. Let us see if there is any $n\in Z$ such that $x=\left( 2m+1 \right)\dfrac{\pi }{6}$. So we have; $$\begin{aligned} & \Rightarrow \dfrac{\pi }{12}+n\dfrac{\pi }{3}=\left( 2m+1 \right)\dfrac{\pi }{6} \\\ & \Rightarrow \dfrac{\pi }{12}+n\dfrac{\pi }{3}=m\dfrac{\pi }{3}+\dfrac{\pi }{6} \\\ & \Rightarrow \left( n-m \right)\dfrac{\pi }{3}=\dfrac{\pi }{6}-\dfrac{\pi }{12} \\\ & \Rightarrow n-m=3\left( \dfrac{2-1}{12} \right)=\dfrac{1}{4} \\\ \end{aligned}$$ Since $n,m$ are integers by closure law $n-m$ have to be an integer. So the above result is a contradiction, there is no $n\in Z$such that $x=\left( 2m+1 \right)\dfrac{\pi }{6}$. We can also use $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $ and the general solution of $\cos \theta =\cos \alpha $ as $\theta =2n\pi \pm \alpha $ to solve.