Question #f5b3e

1 Answer
Sean
Jan 25, 2018

Please see below.

Explanation:

.

If, by intervals, you mean the period of the function we can find it as follows:

#2sin^2(4x)-1=0#

We know there is an identity that says:

#sin^2theta=(1-cos2theta)/2#

Using this identity, we get:

#2((1-cos8x)/2)-1=0#

#1-cos8x-1=0#

#-cos8x=0#

#cos8x=0#

Between #0# and #2pi#, we have the following solutions:

#8x=pi/2, (3pi)/2#

#x=pi/16, (3pi)/16#

The period of the function can be found by dividing the period of a cosine function which is #2pi#:by the coefficient of the angle in the function, which is #8#

#(2pi)/8=pi/4#

Here is the graph of it:

enter image source here

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