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

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In summary, the conversation discusses solving for θ in the equation 2sinθcosθ + 1 - 2sin^2θ = 0. Multiple approaches are suggested, including using half/double angle identities or using the base Pythagorean identity on cosine. The final answer is found to be 67.5 degrees.
  • #1
Vee9
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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)
 
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  • #2
Try using
1- 2 sin^2(x) = cos2x.

Then write cos2x + sin2x as cos(2x + angle).
 
  • #3
You also could write it as [itex]\cos 2x + \sin 2x = 0 \Longrightarrow sin2x = -cos2x \Longrightarrow tan2x = -1[/itex]
That last part should be easy to solve. No factoring needed. ;)
 
Last edited:
  • #4
Bohrok said:
You also could write it as [itex]\cos 2x + \sin 2x = 0 \Longrightarrow sin2x = -cos2x \Longrightarrow tan2x = -1[/itex]

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.
 
  • #5
There are multiple paths.

Apply both half/double angle identities, i.e. also [tex]\sin^2(\theta)=(1-\cos(2\theta))/2[/tex]
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.
 
  • #6
Vee9 said:
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.
 
  • #7
jambaugh said:
There are multiple paths.

Apply both half/double angle identities, i.e. also [tex]\sin^2(\theta)=(1-\cos(2\theta))/2[/tex]
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. :[
 

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