solve 1+ cos2x = 2sin^2x

steller

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solve 1+cos2x=2sin2x\displaystyle 1+ cos2x = 2sin^2x

Okay, so i think i should use the trig identity

cos2x=12sin2x\displaystyle cos2x = 1-2sin^2x

joining them i get:

(12sin2x)=2sin2x\displaystyle (1-2sin^2x) = 2sin^2x

from there i dont know what to do.
 
Applying the identity you used, you would actually get:

1+(12sin2(x))=2sin2(x)\displaystyle 1+\left(1-2\sin^2(x) \right)=2\sin^2(x)

Simplifying you get:

sin2(x)=12\displaystyle \sin^2(x)=\frac{1}{2}

Another way to go, is to use the identity 2sin2(x)=1cos(2x)\displaystyle 2\sin^2(x)=1-\cos(2x) to get:

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

which simplifies to:

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

Can you proceed with either or both methods?
 
Applying the identity you used, you would actually get:


Simplifying you get:

sin2(x)=12\displaystyle \sin^2(x)=\frac{1}{2}

Iam not sure how you got this

Can you proceed with either or both methods?

yes
 
First, remove the parentheses:

1+12sin2(x)=2sin2(x)\displaystyle 1+1-2\sin^2(x)=2\sin^2(x)

22sin2(x)=2sin2(x)\displaystyle 2-2\sin^2(x)=2\sin^2(x)

2=4sin2(x)\displaystyle 2=4\sin^2(x)

24=sin2(x)\displaystyle \dfrac{2}{4}=\sin^2(x)

sin2(x)=12\displaystyle \sin^2(x)=\dfrac{1}{2}
 
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