Question

How do you solve \[3\sin \left( {2x} \right) + \cos \left( {2x} \right) = 0\] from 0 to 2pi?

Answer 1

Hint: This problem deals with solving the problem with the help of trigonometric compound angles formula. Here we assign certain variables to the given equation and then apply the cosine compound triangle, which is given by:
$ \Rightarrow \cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)$

Complete step-by-step answer:
Here consider the given equation as shown below:
$ \Rightarrow 3\sin \left( {2x} \right) + \cos \left( {2x} \right) = 0$
Here considering the left hand side of the above equation, as shown:
Now rewriting the left hand side such that it can be seen as a cosine compound angle.
Let $3 = P\sin C$ and \[1 = P\cos C\]
We know that $P\sin C\sin \left( {2x} \right) + P\cos C\cos \left( {2x} \right) = P\left( {\cos \left( {C - 2x} \right)} \right)$ from cosine compound angle, as simplified below:
$ \Rightarrow P\sin C\sin \left( {2x} \right) + P\cos C\cos \left( {2x} \right) = P\left( {\cos \left( {C - 2x} \right)} \right)$
Here $P = \sqrt {{{\sin }^2}C + {{\cos }^2}C} $
Now finding the value of the $\sqrt {{{\sin }^2}C + {{\cos }^2}C} $ as shown below:
$ \Rightarrow \sqrt {{{\sin }^2}C + {{\cos }^2}C} = \sqrt {{1^2} + {3^2}} $
$\therefore P = \sqrt {10} $
Now substituting this in the expression of obtained cosine compound angle formula, as shown below:
$ \Rightarrow \sqrt {10} \sin C\sin \left( {2x} \right) + \sqrt {10} \cos C\cos \left( {2x} \right) = \sqrt {10} \left( {\cos \left( {C - 2x} \right)} \right)$
Now finding the value of $C$,
Here we assumed that $1 = P\cos C$, from here we are going to obtain the value of $C$, as shown below:
$ \Rightarrow \cos C = \dfrac{1}{P}$
$ \Rightarrow \cos C = \dfrac{1}{{\sqrt {10} }}$
Now applying the inverse cosine to the above equation as shown:
$ \Rightarrow C = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {10} }}} \right)$
$\therefore C = 1.249$
Now substituting the value of $C$, in the expression as shown below:
\[ \Rightarrow 3\sin \left( {2x} \right) + \cos \left( {2x} \right) = \sqrt {10} \cos \left( {1.249 - 2x} \right)\]
Now the right hand side is zero as already given, which is shown below:
\[ \Rightarrow \sqrt {10} \cos \left( {1.249 - 2x} \right) = 0\]
\[ \Rightarrow \cos \left( {2x - 1.249} \right) = 0\]
When the value of cosine is zero, then the value of the angle is a general solution of $\dfrac{\pi }{2} + \pi n$, where n = 0, 1, 2,…
\[ \Rightarrow \left( {2x - 1.249} \right) = {\cos ^{ - 1}}\left( 0 \right)\]
\[ \Rightarrow \left( {2x - 1.249} \right) = \dfrac{\pi }{2} + \pi n\]
Substituting the value of $\pi $, as shown below:
\[ \Rightarrow 2x = 1.249 + \dfrac{{3.14}}{2} + \pi n\]
\[ \Rightarrow 2x = 2.8198 + \pi n\]
The general value of $x$ is given by:
\[\therefore x = 1.4099 + \dfrac{\pi }{2}n\]
Now substituting the values of $n = 0,1,2,3$ as the given interval is from $0$ to $2\pi $.
For $n = 0$
\[ \Rightarrow x = 1.4099\]
For $n = 1$
\[ \Rightarrow x = 2.9807\]
For $n = 2$
\[ \Rightarrow x = 4.5515\]
For $n = 3$
\[ \Rightarrow x = 6.1223\]

$x = \left\{ {1.4099,2.9807,4.5515,6.1223} \right\}$

Note:
Please note that while solving this problem we used cosine trigonometric compound angles, and there are compound angles for sine and tangent as well, which are:
$ \Rightarrow \sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B$
$ \Rightarrow \cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B$
$ \Rightarrow \tan \left( {A \pm B} \right) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$