Solving Trig Equations: sin2θ - 1 = cos2θ | 0 ≤ θ ≤ 360°

[BOAS]

Homework Statement

Solve sin2θ - 1 = cos2θ in range 0 ≤ θ ≤ 360°

Homework Equations

The Attempt at a Solution

I always struggle with the end of these questions, deciding which answers are correct.

Here's what I have done;

let 2θ = x

cosx + 1 = sinx

cos²x + sin²x = 1
sin x = √(1 - cos²x)

cosx + 1 = √(1 - cos²x)

square both sides

cos²x + 2cosx + 1 = 1 - cos²x

2cos²x + 2cosx = 0

factor out 2cosx

2cosx (cosx + 1) = 0

cosx = 0 when x = 270, 90

so θ = 135, 45

cosx - 1 = 0

x = 180, 270

θ = 90, 135

Only 90 and 45 are in the first quadrant where cos is +ve, does that mean these are the answers I want?

Thanks for any help you can give, I hope my question is clear.

 

[adjacent]

cos²x + sin²x = 1
sin x = √(1 - cos²x)

cosx + 1 = √(1 - cos²x)

square both sides

cos²x + 2cosx + 1 = 1 - cos²x

2cos²x + 2cosx = 0

What is cos² and sin²? It does not have any value.

 

[BOAS]

What is cos² and sin²? It does not have any value.

it's cos² of x, which I'm using to represent 2θ

I don't really understand what you're asking

 

[Saitama]

cosx - 1 = 0

x = 180, 270

θ = 90, 135

Recheck this. What are the solutions to cos(x)=1?

 

[BOAS]

Recheck this. What are the solutions to cos(x)=1?

cosx = 1 when x = 0 or 360.

 

[Mark44]

cos²x + sin²x = 1
sin x = √(1 - cos²x)

cosx + 1 = √(1 - cos²x)

square both sides

cos²x + 2cosx + 1 = 1 - cos²x

2cos²x + 2cosx = 0

What is cos² and sin²? It does not have any value.

What's the reason for your comment, adjacent? I don't see anything wrong with the notation that BOAS used.

 

[Saitama]

cosx = 1 when x = 0 or 360.

Since $0 \leq \theta \leq 2\pi$, therefore $0\leq x \leq 4\pi$, you should find solutions for x in this range.

 

[adjacent]

What's the reason for your comment, adjacent? I don't see anything wrong with the notation that BOAS used.

It's because cos itself does not have any value,it's a function.Then how can cos² have any value?

For example, $f(x)=2x+1$
$f$ itself has no value nor $f^2$I know I am wrong somewhere and I would appreciate if you clarify my doubt.

 

[BOAS]

Since $0 \leq \theta \leq 2\pi$, therefore $0\leq x \leq 4\pi$, you should find solutions for x in this range.

this does make sense, but I struggle with the logic of which solutions are the ones I want.

In my OP at the end, I justified the solutions i 'chose' by saying that they were the only ones in a quadrant where cos is +ve.

Is this what I want?

i.e in the range $0\leq x \leq 4\pi$ there are two revolutions to go through, so I think there should be another set of solutions that lie within the first quadrant.

 

[Saitama]

In my OP at the end, I justified the solutions i 'chose' by saying that they were the only ones in a quadrant where cos is +ve.

It can be positive in both the first and fourth quadrants.

 

[BOAS]

No, I'm somewhat confused.

cos(x) = 0

x = 90, 270, 450, 630

so θ = 45, 135, 225, 315

cos(x) = 1

x = 0, 360, 720 (I think I'm correct not to include the -1 values here)

so θ = 0, 180, 360

 

[Saitama]

No, I'm somewhat confused.

cos(x) = 0

x = 90, 270, 450, 630

so θ = 45, 135, 225, 315

cos(x) = 1

x = 0, 360, 720 (I think I'm correct not to include the -1 values here)

so θ = 0, 180, 360

Looks very good to me but it isn't necessary that the values of you have found are the solutions to the original equation.

While squaring, there is always a risk of getting extra solutions so the best method is to find a way which does not involve squaring.

For the given case, you can use:

sin(2θ)=2sinθcosθ and cos(2θ)=2cos²θ-1

Are you aware of the above identities?

 

[BOAS]

Looks very good to me but it isn't necessary that the values of you have found are the solutions to the original equation.

While squaring, there is always a risk of getting extra solutions so the best method is to find a way which does not involve squaring.

For the given case, you can use:

sin(2θ)=2sinθcosθ and cos(2θ)=2cos²θ-1

Are you aware of the above identities?

They're not committed to memory, but they look familiar.

I shall rework the question using them and see how I get on. Thanks for helping me.

 

[Mark44]

It's because cos itself does not have any value,it's a function.Then how can cos² have any value?

For example, $f(x)=2x+1$
$f$ itself has no value nor $f^2$

Are you asking about this? cos²x + sin²x = 1

BOAS write several equations that involved cos²x or sin²x, and these are well-understood forms of notation.

cos²x means exactly the same thing as (cos(x))².

 

[Chestermiller]

If, in your first equation, you move the cosx to the right side of the equation, and square the resulting relationship, you get

cos²x+sin²x-2sinxcosx=1
So, sinxcosx=0. So, at the solution points, either sinx is zero or cosx is zero. You have to check with the original equation to see which such points are allowed.