How to possibly factor this?Solving for θ in a Trigonometric Equation

[Vee9]

Homework Statement

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

The Attempt at a Solution

I replaced 2sinxcoxx with sin2x.
so I ended up with:

-2sin^2(x) + sin2x + 1 = 0

How should I continue to solve for theta? (or x)

 

[╔(σ_σ)╝]

Try using
1- 2 sin^2(x) = cos2x.

Then write cos2x + sin2x as cos(2x + angle).

 

[Bohrok]

You also could write it as $\cos 2x + \sin 2x = 0 \Longrightarrow sin2x = -cos2x \Longrightarrow tan2x = -1$
That last part should be easy to solve. No factoring needed. ;)

 

[Vee9]

You also could write it as $\cos 2x + \sin 2x = 0 \Longrightarrow sin2x = -cos2x \Longrightarrow tan2x = -1$

The answer is 67.5.
So from the tan2x = -1 that you found, I did:
45+ (45/2) = 67.5

One thing I didn't understand was how you got
tan2x = -1 from sin2x = -cos2x ?
Thanks. Major mind block today, lol.

 

[jambaugh]

There are multiple paths.

Apply both half/double angle identities, i.e. also $$\sin^2(\theta)=(1-\cos(2\theta))/2$$
This gives you an equation you can solve directly because of its simple form.

One other approach is to not use half angle identities but to use the base pythagorean identity on the cosine. You'll get a square root so solve for that term and square it out. The result will be a quartic equation but its really a quadratic equation in sine squared of theta so you can solve it easily.

 

[╔(σ_σ)╝]

The answer is 67.5.
So from the tan2x = -1 that you found, I did:
45+ (45/2) = 67.5

One thing I didn't understand was how you got
tan2x = -1 from sin2x = -cos2x ?
Thanks. Major mind block today, lol.

Divide both sides by cos2x.

 

[Vee9]

There are multiple paths.

Apply both half/double angle identities, i.e. also $$\sin^2(\theta)=(1-\cos(2\theta))/2$$
This gives you an equation you can solve directly because of its simple form.

One other approach is to not use half angle identities but to use the base pythagorean identity on the cosine. You'll get a square root so solve for that term and square it out. The result will be a quartic equation but its really a quadratic equation in sine squared of theta so you can solve it easily.

I wish I could do the first method that you said.
Except that we didn't learn the half angle ID's. :[