How to solve this trig equation!

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November 15th 2010, 06:40 PM
TN17

How to solve this trig equation!

1. The problem statement, all variables and given/known data

Solve for θ.
2sinθcosθ + 1 - 2sin^2θ = 0

2. Relevant equations

No equations, just identities.

3. The attempt at a solution
I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
I don't know how to simplify it so that everything is in terms of one similar Identity.
:\
November 15th 2010, 06:55 PM
dwsmith

$\displaystyle 2sinxcosx=sin2x$

Make the sub:

$\displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....$
November 15th 2010, 07:17 PM
TN17

Quote:

Originally Posted by dwsmith

$\displaystyle 2sinxcosx=sin2x$

Make the sub:

$\displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....$

I think you might have misread that I meant -2sin(square)x instead of -2sin2x, lol.

If I subbed that
I would get:
-2sin^2(x) + sin2(x) + 1 = 0
I wanted to factor, but I got stuck at the sin2(x) part. I only know how to factor if the 2 was in front...?
November 15th 2010, 07:31 PM
mr fantastic

Quote:

Originally Posted by TN17

1. The problem statement, all variables and given/known data

Solve for θ.
2sinθcosθ + 1 - 2sin^2θ = 0

2. Relevant equations

No equations, just identities.

3. The attempt at a solution
I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
I don't know how to simplify it so that everything is in terms of one similar Identity.
:\

First substitute $\displaystyle 2 \sin(\theta) \cos(\theta) = \sin(2 \theta)$ and $\displaystyle \sin^2 (\theta) = \frac{1 - \cos(2 \theta)}{2}$ and simplify.