The number of Solutions in #[0,2pi]# such that #(Sin (2x))^4 = 1/8# is?

1 Answer
Oct 4, 2017

Answer:

There are 4 solutions.

Explanation:

Right now, we have: #sin^4(2x)=1/8#

First, we must remove the #""^4# by taking the fourth root of each side:
#root(4)(sin^4(2x))=sin(2x)#

#sin(2x)=root(4)(1/8)=(root(4)(1))/(root(4)(8))=+-1/1.682=+-0.594530321#

Now, we must take the arcsin of each side:
#arcsin(sin(2x))=arcsin(+-0.594530321)-=2x=sin^(-1)(+-0.594530321)#

Since our coefficient of #x# is 2, our range must be doubled to #[0,4\pi]#

#sin^(-1)(0.594530321)=2x=0.20266198\pi and 2.20266198\pi#
#sin^(-1)(-0.594530321)=2x=1.79733802pi and 3.79733802pi\pi#

Therefore, #x=0.10133099\pi, 0.89866901\pi, 1.89866901\pi and 1.10133099\pi#, so there are 4 solutions.