Solve: cos^2 theta/2 - cos theta =0 for 0<or = theta <2Phi

[SueC01]

I hope someone can help. I have been working on a couple of trig problems and can't quite get them. Here is one of them:

Solve: cos^2 theta/2 - cos theta =0 for 0<or = theta <2Phi

The answer I came up with is 0 and I am not even sure of that.

Could you please show me the equation used for the answer.

I am so ready to pull my hair out.

Thank you,
Sue

 

[Deleted member 4993]

Re: Help with Trig please.

I hope someone can help. I have been working on a couple of trig problems and can't quite get them. Here is one of them:

Solve: cos^2 theta/2 - cos theta =0 for 0<or = theta <2Phi

Solution to your problem:

\(\displaystyle \cos^2(\frac{\theta}{2}) - \cos(\theta) \, = \, 0 \, for \, 0 \, \le \theta < 2\pi\)

is correct

The answer I came up with is 0 and I am not even sure of that.

Could you please show me the equation used for the answer.

I am so ready to pull my hair out.

Thank you,
Sue

 

[soroban]

Re: Help with Trig please.

Hello, Sue!

\(\displaystyle \text{Solve: }\:\cos^2\!\tfrac{\theta}{2} - \cos\theta \:=\:0\quad\text{for }0\leq \theta < 2\pi\)

The answer I came up with is 0 and I am not even sure of that.
Of course, it depends on HOW you came up with it . . .

\(\displaystyle \text{I would use the identity: }\:\cos^2\!\tfrac{\theta}{2} \:=\:\frac{1 + \cos\theta}{2}\)

\(\displaystyle \text{Then we have: }\:\frac{1 + \cos\theta}{2} - \cos\theta \:=\:0 \quad\Rightarrow\quad \cos\theta \:=\:1 \quad\Rightarrow\quad \theta \:=\:0\)

 

[SueC01]

Re: Help with Trig please.

Thank you both very very much. After going to another site and getting totally insulted I was ready to give up on asking for any help.

Again,
Thank you, very much,
Sue