Trig Identities: Solving Equations

[Monkeyseat]

If someone could just help me get to form a "quadratic" type equation with one variable for these questions, I can take it from there.

I am stuck on part (d) and (e):

Like stated in my other thread, I think I have to use this:

But I can't seem to get one variable to form an equation for these 2.

Any ideas?

Thanks.

EDIT:

Is this right for (e)? I'm just going to swap theta for x because it's easier to type:

tanx * sinx - cosx = 0
tanx * sinx = cos x
tanx * sinx = (sinx)/(tanx)
tan^2 x * sinx = sin x
tan^2 x = 1

 

[tkhunny]

Start each by getting rid of the tangents.

tan(x) + cos(x) = sin(x)/cos(x) + cos(x) = 0

 

[Monkeyseat]

Start each by getting rid of the tangents.

tan(x) + cos(x) = sin(x)/cos(x) + cos(x) = 0

I was thinking about that. Any chance of another suggestion... :?

For d:

sinx/cosx + cosx = 0
sinx/cosx = -cosx
sinx = -cos^2 x

cos^2x = 1 - sin^2x

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

Is that ok?

I did this for e:

tanx * sinx - cosx = 0
tanx * sinx = cos x
tanx * sinx = (sinx)/(tanx)
tan^2 x * sinx = sin x
tan^2 x = 1

I know you said eliminate tangents but I did this beforehand, is that incorrect? Not sure how to do what you said on (e)...

 

[Monkeyseat]

Ok - after working hard I've done both of these now I think. :mrgreen:

For (e) I did:

(sinx/cosx) * sinx - cosx = 0

And after changing it around a bit:

2sin^2 x - 1 = 0

Many thanks for your help! By the way, was the "alternate" method I posted in my above post also correct that was arranged for tangent?

 

[tkhunny]

You do want to avoid Domain issues. Turning cos(x) into sin(x)/tan(x) is a little dangerous. See if this works for \(\displaystyle x=\frac{\pi}{2}\).

 

[Monkeyseat]

You do want to avoid Domain issues. Turning cos(x) into sin(x)/tan(x) is a little dangerous. See if this works for \(\displaystyle x=\frac{\pi}{2}\).

Ok I'm not quite up to that yet, I'll leave it with the method you suggested and 2sin^2 x - 1 = 0 then if that's correct.

Thanks again.