solve 1+ cos2x = 2sin^2x

[steller]

solve \(\displaystyle 1+ cos2x = 2sin^2x \)

Okay, so i think i should use the trig identity

\(\displaystyle cos2x = 1-2sin^2x \)

joining them i get:

\(\displaystyle (1-2sin^2x) = 2sin^2x \)

from there i dont know what to do.

 

[MarkFL]

Applying the identity you used, you would actually get:

\(\displaystyle 1+\left(1-2\sin^2(x) \right)=2\sin^2(x)\)

Simplifying you get:

\(\displaystyle \sin^2(x)=\frac{1}{2}\)

Another way to go, is to use the identity \(\displaystyle 2\sin^2(x)=1-\cos(2x)\) to get:

\(\displaystyle 1+\cos(2x)=1-\cos(2x)\)

which simplifies to:

\(\displaystyle \cos(2x)=0\)

Can you proceed with either or both methods?

 

[steller]

Applying the identity you used, you would actually get:


Simplifying you get:

\(\displaystyle \sin^2(x)=\frac{1}{2}\)

Iam not sure how you got this

Can you proceed with either or both methods?

yes

 

[MarkFL]

First, remove the parentheses:

\(\displaystyle 1+1-2\sin^2(x)=2\sin^2(x)\)

\(\displaystyle 2-2\sin^2(x)=2\sin^2(x)\)

\(\displaystyle 2=4\sin^2(x)\)

\(\displaystyle \dfrac{2}{4}=\sin^2(x)\)

\(\displaystyle \sin^2(x)=\dfrac{1}{2}\)