Question

How do you solve the $3\sin \left( 2x \right)+\cos \left( 2x \right)=0$ from $0$ to $2\pi $?

Answer 1

Hint: From the given question we have to solve the $3\sin \left( 2x \right)+\cos \left( 2x \right)=0$ in the domain of $0$to $2\pi $. To solve the above question, we have to use the trigonometric concepts like $A\cos x+B\sin x=C\cos \left( x-D \right)$ where $C=\sqrt{{{A}^{2}}+{{B}^{2}}}$, $\cos D=\dfrac{A}{C}$ and $\sin D=\dfrac{B}{C}$. By this we can solve the question by using this transformation we will get solutions from $0$to $2\pi $.

Complete step by step solution:
From the question given we have to solve the $3\sin \left( 2x \right)+\cos \left( 2x \right)=0$ from $0$ to $2\pi $.
$\Rightarrow 3\sin \left( 2x \right)+\cos \left( 2x \right)=0$
We have to use the trigonometric transformation in the given question.
The transformation we have to use in this question is
$\Rightarrow A\cos x+B\sin x=C\cos \left( x-D \right)$
In this the values of C and D are,
$\Rightarrow C=\sqrt{{{A}^{2}}+{{B}^{2}}}$
Where as $\cos D$ and $\sin D$ are
$\Rightarrow \cos D=\dfrac{A}{C}$
$\Rightarrow \sin D=\dfrac{B}{C}$
Now comparing the transformation with the given equation, we will get,
$\Rightarrow A=1,\text{ B=3}$
From this we will get the values of C and D are,
$\Rightarrow \text{C=}\sqrt{{{1}^{2}}+{{3}^{2}}}=\sqrt{10}$
 Where as $\cos D$ and $\sin D$ are
$\Rightarrow \cos D=\dfrac{1}{\sqrt{10}}$
$\Rightarrow \sin D=\dfrac{3}{\sqrt{10}}$
As from the transformation we only need the value of $\cos D$,
From this we will get the value of D,
$\Rightarrow D={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right)=1.249\ldots $
From the transformation we can write the above expression, which is given in the question is,
$\Rightarrow 3\sin \left( 2x \right)+\cos \left( 2x \right)=\sqrt{10}\cos \left( 2x-1.249\ldots \right)=0$
Now by using this we will get
$\Rightarrow \sqrt{10}\cos \left( 2x-1.249\ldots \right)=0$
$\Rightarrow \cos \left( 2x-1.249\ldots \right)=0$
$\Rightarrow 2x-1.249\ldots ={{\cos }^{-1}}\left( 0 \right)$
$\Rightarrow 2x-1.249\ldots =\dfrac{\pi }{2}+\pi n$
$\Rightarrow 2x=1.249+\dfrac{\pi }{2}+\pi n$
$\Rightarrow 2x=2.8198\ldots +\pi n$
$\Rightarrow x=1.4099\ldots +\dfrac{\pi }{2}n$
Now we got the general solution by putting the values of n we will get
First, we have to put $n=0$
$\Rightarrow n=0,x=1.4099$
Now we have to put $n=1$
$\Rightarrow n=1,x=2.9807$
Now we have to put $n=2$
$\Rightarrow n=2,x=4.5515$
Now we have to put $n=3$
$\Rightarrow n=3,x=6.1223$
Now we have to put $n=4$
$\Rightarrow n=4,x=7.6931$
Therefore, $7.6931>2\pi $ we have to stop because we need from $0$ to $2\pi $
Therefore, the solution set S is
$\Rightarrow S=\left\{ 1.4099,2.9807,4.5515,6.1223 \right\}$

Note: Student should know the trigonometric transformations and here $\sqrt{10}\cos \left( 2x-1.249\ldots \right)=0$ students should make sure that to carry all the digits until the end then round off. Students should know that cosine and sine are both positive means quadrant one. Students should be careful while doing calculations.