Question

Find the general solution of $\cos 3x = \sin 2x$ .

Answer 1

Hint: Equations involving trigonometric functions of a variable are called trigonometric equations.
The value of the variable (or unknown) which satisfies the trigonometric equations is called solutions.
We know that the values of $\sin x$ and $\cos x$ repeat after an interval of $2\pi $ and the values of $\tan x$ repeat after an interval of $\pi $.
The solutions of a trigonometric equation for which $0 \leqslant x \leqslant 2\pi $ are called principal solutions.
The expression involving integer $'n'$ which gives all solutions of a trigonometric equation is called the general solution.
Always factorize the equation to find the variables.
For $\sin x = 0$gives $x = n\pi $, where $n \in \mathbb{Z}$.
For $\cos x = 0$ gives $x = \left( {2n + 1} \right)\dfrac{\pi }{2}$ , where $n \in \mathbb{Z}$.
In related questions, you might come across the situation like:
$\sin x = \sin \dfrac{\pi }{3}$;
Use the following theorem to simplify:
For any real numbers x and y,
\[\sin x = \sin y \Rightarrow x = n\pi + {\left( { - 1} \right)^n}y,\] where $n \in \mathbb{Z}$.
For $\sin x = \sin \dfrac{\pi }{3}$
$ \Rightarrow x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{3}$ , where $n \in \mathbb{Z}$.

Complete step-by-step answer:
Step 1: Simplify the given equation:
$\cos 3x = \sin 2x$
Using trigonometric identity:
$
  \cos 3x = 4{\cos ^3}x - 3\cos x \\
  \sin 2x = 2\sin x\cos x \\
 $
The equation becomes:
$
   \Rightarrow {\text{ }}4{\cos ^3}x - 3\cos x = 2\sin x\cos x \\
   \Rightarrow {\text{ }}4{\cos ^3}x - 3\cos x - 2\sin x\cos x = 0 \\
 $
Taking $\cos x$as common
$ \Rightarrow {\text{ }}\cos x\left( {4{{\cos }^2}x - 3 - 2\sin x} \right) = 0$
Using trigonometric identity:
${\cos ^2}x + {\sin ^2}x = 1$
$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
The equation becomes:
\[
   \Rightarrow {\text{ }}\cos x\left[ {4\left( {1 - {{\sin }^2}x} \right) - 3 - 2\sin x} \right] = 0 \\
   \Rightarrow {\text{ }}\cos x\left[ {4 - 4{{\sin }^2}x - 3 - 2\sin x} \right] = 0 \\
 \]
\[ \Rightarrow {\text{ }}\cos x\left( {1 - 4{{\sin }^2}x - 2\sin x} \right) = 0\] ….. (1)
Step 2: Find the general solutions:
If the product of two numbers is 0, then either the first number or second number of both of them are 0.
Thus, $\cos x = 0$ and/or \[1 - 4{\sin ^2}x - 2\sin x = 0\]
For $\cos x = 0$

Graph: $\cos x$
From the graph we can see that $\cos x = 0$for the odd multiple of $\dfrac{\pi }{2}$
$ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{2}$ …… (2)
Where $n \in \mathbb{Z}$ . $\mathbb{Z}$is a set of integers.
For n = 0, $x = \dfrac{\pi }{2}$ .
For n = 1, $x = \dfrac{{3\pi }}{2}$
For \[1 - 4{\sin ^2}x - 2\sin x = 0\]
Let \[\sin x = y\]
Therefore, $1 - 4{y^2} - 2y = 0$
Use the discrimination method to solve the quadratic equation in y.
$
  a{y^2} + by + c = 0 \\
   \Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
 $
$4{y^2} + 2y - 1 = 0$
on comparing with standard quadratic equation:
$
  a = 4;b = 2;c = - 1 \\
   \Rightarrow y = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2\left( 4 \right)}} \\
   \Rightarrow y = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8} \\
   \Rightarrow y = \dfrac{{ - 2 \pm \sqrt {20} }}{8} \\
   \Rightarrow y = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8} \\
   \Rightarrow y = \dfrac{{ - 1 \pm \sqrt 5 }}{2} \\
 $
Thus, $\sin x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$

Graph: $\sin x$
Range of $y = \sin x = \left[ { - 1,1} \right]$
\[\dfrac{{ - 1 - \sqrt 5 }}{2} \simeq - 1.6 < - 1\]
$ \Rightarrow \sin x \ne \dfrac{{ - 1 - \sqrt 5 }}{2}$
So, $\sin x = \dfrac{{ - 1 + \sqrt 5 }}{2}$
We know, \[\sin 18^\circ = \dfrac{{ - 1 + \sqrt 5 }}{2}\]
$ \Rightarrow \sin x = \sin 18^\circ $
To convert degree into radian:$x = \dfrac{\pi }{{10}}$
We know, $180^\circ = \pi $ radians
Therefore, $1^\circ = \dfrac{\pi }{{180^\circ }}$
So, \[18^\circ = 18^\circ \times \dfrac{\pi }{{180^\circ }}\]
$ \Rightarrow {\text{ }} = {\text{ }}\dfrac{\pi }{{10}}$ radians
Or $\sin x = \sin \dfrac{\pi }{{10}}$
We know, \[\sin x = \sin y \Rightarrow x = n\pi + {\left( { - 1} \right)^n}y,\] where $n \in \mathbb{Z}$. $\mathbb{Z}$is a set of integers.
On comparing, $y = \dfrac{\pi }{{10}}$
$ \Rightarrow x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{{10}}$ …… (3)
For n = 0,
For n = 1, $x = \dfrac{{9\pi }}{{10}}$
For n =2, $x = \dfrac{{21\pi }}{{10}}$
The general solution of the equation (1) is the union of equation (2) and (3) as both the value of x satisfies the equation (1)
Final answer: Thus, the general solution of $\cos 3x = \sin 2x$ is $x = \left[ {\left\{ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right\} \cup \left\{ {n\pi + {{\left( { - 1} \right)}^n}\dfrac{\pi }{{10}}} \right\}} \right]$ , where $n \in \mathbb{Z}$.

Note: Similarly for any real numbers x and y,
$\cos x = \cos y \Rightarrow x = 2n\pi \pm y,$ where $n \in \mathbb{Z}$.
$\tan x = \tan y \Rightarrow x = n\pi + y$ , where $n \in \mathbb{Z}$.
Learn trigonometric identities to solve related questions easily.