Question

How do you solve sin4x + 2sin2x = 0?

Answer 1

Hint: Here in this question, first we have to solve the equation by applying some trigonometric identities i.e.
$\Rightarrow $Sin(x + y) = sinx.cosy + cosx.siny
$\Rightarrow $sin2x = 2sinxcosx
Majorly these two formulae will be used. Apart from this, we have found the general solution of the equation after solving it.

Complete answer:
Let’s solve the question now.
Can we just have a look at some basic trigonometric identities.
$\Rightarrow $Cos(-x) = cos(x)
$\Rightarrow $sin(-x) = -sin(x)
$\Rightarrow $cos(x + y) = cosx.cosy – sinx.siny
$\Rightarrow $sin(x + y) = sinx.cosy + cosx.siny
$\Rightarrow $ cos(x - y) = cosx.cosy + sinx.siny
$\Rightarrow $sin(x - y) = sinx.cosy - cosx.siny
$\Rightarrow $cos2x = ${{\cos }^{2}}x-{{\sin }^{2}}x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$
$\Rightarrow $sin2x = 2sinxcosx = $\dfrac{2\tan x}{1+{{\tan }^{2}}x}$
$\Rightarrow $sin3x = 3sinx – $4{{\sin }^{3}}x$
$\Rightarrow $cos3x = $4{{\cos }^{3}}x$- 3cosx
Let’s study general solutions for some basic trigonometric equations:
If sinx = 0, then the general solution for x = $n\pi $
If cosx = 0, then the general solution for x = $n\pi +\dfrac{\pi }{2}$
If sinx = 1, then the general solution for x = $2n\pi +\dfrac{\pi }{2}=\left( 4n+1 \right)\dfrac{\pi }{2}$
If cosx = 1, then the general solution for x = $2n\pi $
Now, write the given equation.
$\Rightarrow $sin4x + 2sin2x = 0
Split the angle 4x into 2x and 2x, we will get:
$\Rightarrow $sin(2x + 2x) + 2sin2x = 0
According to the sum identity:
sin(x + y) = sinx.cosy + cosx.siny
Apply this in equation:
$\Rightarrow $sin2x.cos2x + cos2x.sin2x +2sin2x = 0
Adding the like terms, we get:
$\Rightarrow $2sin2xcos2x +2sin2x = 0
Take 2sin2x common from both the terms:
$\Rightarrow $2sin2x(cos2x + 1) = 0
Now, we got two cases for which we have to find the general solution:
Case 1: 2sin2x = 0
Case 2: cos2x + 1 = 0
For case 1:
$\Rightarrow $2sin2x = 0
Use identity sin2x = 2sinxcosx above:
$\Rightarrow $2(2sinxcosx) = 0
Open bracket and multiply:
$\Rightarrow $4sinxcosx = 0
Take 4 on the other side, we get:
$\Rightarrow $sinxcosx = 0
Whenever sinx = 0 and cosx = 0 then the whole equation will become 0. If sinx = 0, then the general solution for x = $n\pi $.
 For that the value of x will be:
x = $0+2\pi n,\dfrac{\pi }{2}+2\pi n,\pi +2\pi n$
Now, for case 2:
$\Rightarrow $cos2x + 1 = 0
As we know that cos2x = $2{{\cos }^{2}}x-1$. Put this value above:
$\Rightarrow 2{{\cos }^{2}}x-1$ + 1 = 0
Solve the like terms, we get:
$\Rightarrow 2{{\cos }^{2}}x$ = 0
Take 2 on the other side:
$\Rightarrow {{\cos }^{2}}x$ = 0
Take root on both the sides, we get:
$\Rightarrow \cos x$ = 0
If cosx = 0, then the general solution for x = $n\pi +\dfrac{\pi }{2}$
Now for x, general solution will be:
$\Rightarrow x=\dfrac{\pi }{2}+2\pi n$ and $\dfrac{3\pi }{2}+2\pi n$

Note: Important points to be noted are that the equations of trigonometric functions in variable ‘x’, where x lies between $0\le x\le 2\pi $ is called the principal solution. And the equation containing integer ‘n’ in it will be called a general solution. If a question comes in the form of a trigonometric equation, then we have to find the principal or general solution depending upon the conditions.