Question #6192d

1 Answer
Jun 2, 2017

#theta = pi pm pi n#; #n in ZZ#

Explanation:

I'm assuming you mean to solve #(sin^(2)(theta))^(2) = 0#.

#Rightarrow (sin^(2)(theta))^(2) = 0#

#Rightarrow sin^(2)(theta) = 0#

#Rightarrow sin(theta) = 0#

Now, let's take a look at the graph of #y = sin(theta)#:

graph{sin(x) [-10, 10, -5, 5]}

The #x#-axis is the axis for all values of #theta#.

We are looking for the values of #theta# for which #sin(theta) = 0#.

On the graph, we can see that #y = 0# for multiple values of #theta#, so there will be multiple solutions to the equation #sin(theta) = 0#.

Also, you haven't specified a domain, so the solutions can be positive or negative.

enter image source here

The coordinates of some of the solutions are shown in the graph above.

The solutions are all multiples of #pi#, i.e. #theta = pi pm pi n# for some integer #n#.

Let's express the solutions formally:

#therefore theta = pi pm pi n#; #n in ZZ#