Question

sin^2(2 theta)=1 on the interval 0< theta< 2pi

Answer 1

Well if we set \[\large 2\theta\]=a
We get
sin^2(a)=1

We subtract 1 from both sides to get

\[\Large \sin^2(a)-1=0\]

Can you factor this for me?

Answer 2

\[\sin ^{2}(a)=1 \]
\[\sqrt{\sin ^{2}}(a)=\sqrt{1}\]
\[\sin (a)=1\]
\[\sin\frac{ \pi }{ 2 }= 1\]
im not to sure if this is right

Answer 3

You have to factor it out since there's a sqrt

Answer 4

\[\sin^2(a)-1=0~~~~~~~~~~~~~~(\sin(a)+1)*(\sin(a)-1)=0\]

Answer 5

set those those two equations eqaul to 0