1. ## Trignometric equation help.

Solve each equation on the interval 0<= θ < 2n

1) sin² θ - cos² θ = 1 + cos θ

2) cos (2θ) + 6 sin² θ = 4

2. Originally Posted by cloudzer0
Solve each equation on the interval 0<= θ < 2n

1) sin² θ - cos² θ = 1 + cos θ

2) cos (2θ) + 6 sin² θ = 4

1) Replace $\sin^2 \theta$ with $1 - \cos^2 \theta$ and re-arrange to get a simple quadratic in $\cos \theta$:

$\cos \theta (2 \cos \theta + 1) = 0$.

Now solve $\cos \theta = 0$ and $2 \cos \theta + 1 = 0$.

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2) Replace $\cos (2\theta)$ with $1 - 2 \sin^2 \theta$ and re-arrange to get a simple quadratic in $\sin \theta$:

$\sin^2 \theta = \frac{3}{4}$.

Now solve $\sin \theta = \frac{\sqrt3}{2}$ and $\sin \theta = -\frac{\sqrt{3}}{2}$.

3. Thank you for your help, I actually worked it through and figured out those problems myself. However, there are still some confusions going on which I've been looking at for a good while.

Couple of other different problems.

Problem:
cos θ = sin θ

My work:
cos θ - sin θ = 0
=> (cosθ - sin θ)² = 0
=> cos²θ - 2cosθsinθ + sin²θ =0
=> 1-2cosθsinθ = 0

I'm pretty much stuck at this point. Did I do something wrong?

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Another problem which I seem to have trouble with is:
(tan θ - 1)(sec θ - 1) = 0

4. Originally Posted by cloudzer0
Thank you for your help, I actually worked it through and figured out those problems myself. However, there are still some confusions going on which I've been looking at for a good while.

Couple of other different problems.

Problem:
cos θ = sin θ

Mr F says: Divide both sides by $\cos \theta: \,$ $\tan \theta = 0$ .....

My work:
cos θ - sin θ = 0
=> (cosθ - sin θ)² = 0
=> cos²θ - 2cosθsinθ + sin²θ =0
=> 1-2cosθsinθ = 0

I'm pretty much stuck at this point. Did I do something wrong?

*************
Another problem which I seem to have trouble with is:
(tan θ - 1)(sec θ - 1) = 0

Mr F says: Solve the two equations $\tan \theta -1 = 0 \Rightarrow \tan \theta = 1$ ...... and $\sec \theta - 1 = 0 \Rightarrow \frac{1}{\cos \theta} - 1 = 0 \Rightarrow \frac{1}{\cos \theta} = 1 \Rightarrow \cos \theta = 1$ .......
..

5. 1) cos (4θ) - cos (6θ) = 0

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Find real zeros. interval 0<= x < 2n

2) f(x) = 4 cos² x - 1

6. Originally Posted by cloudzer0
1) cos (4θ) - cos (6θ) = 0

Mr F says: $\Rightarrow \cos (4 \theta) = \cos (6 \theta)$.

Now use the fact that the solutions to $\cos \alpha = \cos \beta$ are $\alpha = \beta + 2 n \pi \,$ and $\alpha = - \beta + 2n \pi$ where $n$ is an integer .......

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Find real zeros. interval 0<= x < 2n

2) f(x) = 4 cos² x - 1

Mr F says: Solve $0 = \cos^2 x - 1$ over the given domain.
..