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Question: How do you solve \(\sin 4x - 2\sin 2x = 0\) in the interval \(\left[ {0,360} \right]\)?...

How do you solve sin4x2sin2x=0\sin 4x - 2\sin 2x = 0 in the interval [0,360]\left[ {0,360} \right]?

Explanation

Solution

This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of (A+B)\left( {A + B} \right) or (AB)\left( {A - B} \right)in terms of trigonometric functions of AA and BB. The used formula here is:
sin(A+B)=sinAcosB+cosAsinB\Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B

Complete step-by-step solution:
Given an equation of trigonometric expression functions.
The given equation is sin4x2sin2x=0\sin 4x - 2\sin 2x = 0, consider this as given below:
sin4x2sin2x=0\Rightarrow \sin 4x - 2\sin 2x = 0
We know that by using one of the trigonometric identity, as given below:
Consider only sin4x\sin 4x, applying the sine trigonometric compound angle to it, as given below:
sin4x=sin(2x+2x)\Rightarrow \sin 4x = \sin \left( {2x + 2x} \right)
sin4x=sin2xcos2x+cos2xsin2x\Rightarrow \sin 4x = \sin 2x\cos 2x + \cos 2x\sin 2x
On further simplifying the above expression, as given below:
sin4x=2sin2xcos2x\Rightarrow \sin 4x = 2\sin 2x\cos 2x
Substitute this expression in the given equation of sin4x2sin2x=0\sin 4x - 2\sin 2x = 0, as given below:
2sin2xcos2x2sin2x=0\Rightarrow 2\sin 2x\cos 2x - 2\sin 2x = 0
From the above expression taking the term 2sin2x2\sin 2x common, as given below:
2sin2x(cos2x1)=0\Rightarrow 2\sin 2x\left( {\cos 2x - 1} \right) = 0
Now 2sin2x2\sin 2x should be zero, or cos2x1\cos 2x - 1 should be equal to zero, so considering both the cases, as shown below:
First consider sin2x\sin 2x is equated to zero, as given below:
sin2x=0\Rightarrow \sin 2x = 0
2x=0,180,360,540,720\Rightarrow 2x = 0,180,360,540,720
To get the value of xx, dividing the above expression by 2, as shown below:
x=0,1802,3602,5402,7202\Rightarrow x = 0,\dfrac{{180}}{2},\dfrac{{360}}{2},\dfrac{{540}}{2},\dfrac{{720}}{2}
x=0,90,180,270,360\therefore x = 0,90,180,270,360
Now consider cos2x1\cos 2x - 1 is equated to zero, as given below:
cos2x1=0\Rightarrow \cos 2x - 1 = 0
cos2x=1\Rightarrow \cos 2x = 1
2x=0,360,720\Rightarrow 2x = 0,360,720
To get the value of xx, dividing the above expression by 2, as shown below:
x=0,3602,7202\Rightarrow x = 0,\dfrac{{360}}{2},\dfrac{{720}}{2}
x=0,180,360\therefore x = 0,180,360
Now combining the values of xx, from both the cases, as shown below:
\Rightarrow x = \left\\{ {0,90,180,270,360} \right\\} \cup \left\\{ {0,180,360} \right\\}
\therefore x = \left\\{ {0,90,180,270,360} \right\\}

The values of x in the interval are \left\\{ {0,90,180,270,360} \right\\}.

Note: Please note that while solving the values of xx, in both the cases where sin2x=0\sin 2x = 0 and cos2x=1\cos 2x = 1, here the solutions for xx can be infinite number of solutions in both the cases, but here we are restricted with the given interval [0,360]\left[ {0,360} \right], hence the solutions for xx, are limited to the given interval.