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put this solution on YOUR website!-\sin^2\left(\theta\right)=0 )
Start with the given equation
Let
. So this means that
-\sin^2\left(2u\right)=0 )
Replace
with u and
with 2u
-\left(\sin\left(2u\right)\right)^2=0 )
Rewrite
)
as
\right)^2)
-\left(2\sin\left(u\right)\cos\left(u\right)\right)^2=0 )
Replace
)
with
\cos\left(u\right))
-4\sin^2\left(u\right)\cos^2\left(u\right)=0 )
Square
\left(1-4\cos^2\left(u\right)\right)=0 )
Factor out
)
Now use the zero product property:
=0)
...or...
=0 )
Now let's solve
:
=0 )
=0 )
Take the square root of both sides
...or...
Take the arcsine of both sides
Since
, this means
...or...
However, since
is not in the interval [0,2
), it is not a solution.
So the first solution is
--------------------------------------------------------------------
Now let's solve
=0 )
:
=-1 )
Subtract 1 from both sides
=\frac{1}{4} )
Divide both sides by -4
Take the square root of both sides:
=\frac{1}{2})
...or...
=-\frac{1}{2} )
So let's solve the first part
=\frac{1}{2} )
:
...or...
...or...
However since our interval is positive, the negative answer is not in the interval.
So another part of the solution is
---------------------
Now let's solve the second part
:
...or...
...or...
So another part of the solution is
===============================================
Answer:
So all together our solutions are:
(