Question

Solve the trigonometric equation \[\sin 3x+\cos 2x=0\].

Answer 1

Hint: In this type of question we have to use the concept of trigonometry. Here, we have to use different formulae of trigonometry such as \[\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \], \[\cos \left( \pi -\theta \right)=-\cos \theta \] and \[\cos \theta =\cos \left( 2n\pi \pm \theta \right)\]. By using these trigonometry formulae we can simplify the given expression to obtain the general solution i.e. to obtain the value of \[x\].

Complete step-by-step solution:
Now, we have to solve \[\sin 3x+\cos 2x=0\].
For this, let us consider,
\[\Rightarrow \sin 3x+\cos 2x=0\]
\[\Rightarrow \cos 2x=-\sin 3x\]
Now, as we know that, \[\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \] we can rewrite the above equation as,
\[\Rightarrow \cos 2x=-\cos \left( \dfrac{\pi }{2}-3x \right)\]
Here, we can observe that our right hand side is negative of cosine so to make it positive, we will use \[\cos \left( \pi -\theta \right)=-\cos \theta \] and hence, we get,
\[\Rightarrow \cos 2x=\cos \left( \pi -\left( \dfrac{\pi }{2}-3x \right) \right)\]
On simplifying the angle present on right hand side, we get,
\[\Rightarrow \cos 2x=\cos \left( \dfrac{\pi }{2}+3x \right)\]
We know that, \[\cos \theta =\cos \left( 2n\pi \pm \theta \right)\] hence, we can write,
\[\Rightarrow \cos 2x=\cos \left( 2n\pi \pm \left( \dfrac{\pi }{2}+3x \right) \right)\]
\[\Rightarrow 2x=\left( 2n\pi \pm \left( \dfrac{\pi }{2}+3x \right) \right)\]
So, we can rewrite the above expression as
\[\Rightarrow 2x=2n\pi +\dfrac{\pi }{2}+3x\text{ or }2x=2n\pi -\dfrac{\pi }{2}-3x\]
By simplifying this further we get,
\[\begin{align}
  & \Rightarrow 2x-3x=2n\pi +\dfrac{\pi }{2}\text{ or }2x+3x=2n\pi -\dfrac{\pi }{2} \\
 & \Rightarrow -x=2n\pi +\dfrac{\pi }{2}\text{ or 5}x=2n\pi -\dfrac{\pi }{2} \\
 & \Rightarrow x=-2n\pi -\dfrac{\pi }{2}\text{ or 5}x=2n\pi -\dfrac{\pi }{2} \\
\end{align}\]
After simplifying the equation further we get,
\[\begin{align}
  & \Rightarrow x=-\dfrac{\pi }{2}\left( 4n+1 \right)\text{ or 5}x=\dfrac{\pi }{2}\left( 4n-1 \right) \\
 & \Rightarrow x=-\dfrac{\pi }{2}\left( 4n+1 \right)\text{ or }x=\dfrac{\pi }{10}\left( 4n-1 \right) \\
\end{align}\]
Hence, the general solution is given by,
\[\Rightarrow x=-\dfrac{\pi }{2}\left( 4n+1 \right)\text{ or }\dfrac{\pi }{10}\left( 4n-1 \right)\text{ where }n\in \mathbb{Z}\]
Thus, on solving the equation \[\sin 3x+\cos 2x=0\] we get its general solution as \[x=-\dfrac{\pi }{2}\left( 4n+1 \right)\text{ or }\dfrac{\pi }{10}\left( 4n-1 \right)\text{ where }n\in \mathbb{Z}\].

Note: In this type of question students may make mistakes after the step \[\cos 2x=\cos \left( \dfrac{\pi }{2}+3x \right)\]. Students have to take care after this step as direct cancellation will lead to a particular answer that will be \[x=-\dfrac{\pi }{2}\] and as we have to solve the given equation for all possible values of \[x\] generalisation is important. So that students have to remember to use the formula \[\cos \theta =\cos \left( 2n\pi \pm \theta \right)\] which will generalise our final answer in the form of \[n\] where \[n\in \mathbb{Z}\]