Question

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

Answer 1

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,

& \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{aligned}$$ After simplifying the equation further we get, $$\begin{aligned} & \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{aligned}$$ Hence, the general solution is given by, $$\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}$$