Find the exact solution of the given equation in the interval [0,2∏)

Question

Find the exact solution of the given equation in the interval [0,2∏)

sin 4x=-2sin2x

Answer 1

sin4x = (2(2x) = 2sin2xcos2x = -2sin2x

cancel the 2sin2x's leaving only

cos2x =-1

2x = pi

x=pi/2

Answer 2

We are going to use the trigonometric identity sin(2a) = 2sin(a)cos(a) such that

sin(4x) = sin(2*(2x)) = 2sin(2x)cos(2x)

Now, since we were given that sin(4x) = -2sin(2x), we can use the result obtained above such that

-2sin(2x) = 2sin(2x)cos(2x)

-2 = 2cos(2x)

-1 = cos(2x)

We can clearly see that 2x = pi, or

x = pi/2