Question
Question: How do you solve \(\sin 4x - 2\sin 2x = 0\) in the interval \(\left[ {0,360} \right]\)?...
How do you solve sin4x−2sin2x=0 in the interval [0,360]?
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) or (A−B)in terms of trigonometric functions of A and B. The used formula here is:
⇒sin(A+B)=sinAcosB+cosAsinB
Complete step-by-step solution:
Given an equation of trigonometric expression functions.
The given equation is sin4x−2sin2x=0, consider this as given below:
⇒sin4x−2sin2x=0
We know that by using one of the trigonometric identity, as given below:
Consider only sin4x, applying the sine trigonometric compound angle to it, as given below:
⇒sin4x=sin(2x+2x)
⇒sin4x=sin2xcos2x+cos2xsin2x
On further simplifying the above expression, as given below:
⇒sin4x=2sin2xcos2x
Substitute this expression in the given equation of sin4x−2sin2x=0, as given below:
⇒2sin2xcos2x−2sin2x=0
From the above expression taking the term 2sin2x common, as given below:
⇒2sin2x(cos2x−1)=0
Now 2sin2x should be zero, or cos2x−1 should be equal to zero, so considering both the cases, as shown below:
First consider sin2x is equated to zero, as given below:
⇒sin2x=0
⇒2x=0,180,360,540,720
To get the value of x, dividing the above expression by 2, as shown below:
⇒x=0,2180,2360,2540,2720
∴x=0,90,180,270,360
Now consider cos2x−1 is equated to zero, as given below:
⇒cos2x−1=0
⇒cos2x=1
⇒2x=0,360,720
To get the value of x, dividing the above expression by 2, as shown below:
⇒x=0,2360,2720
∴x=0,180,360
Now combining the values of x, 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 x, in both the cases where sin2x=0 and cos2x=1, here the solutions for x can be infinite number of solutions in both the cases, but here we are restricted with the given interval [0,360], hence the solutions for x, are limited to the given interval.