Solve:

- Jamierox4ev3r

Solve:

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- Jamierox4ev3r

Give me a second to type the equation, thanks :)

- Jamierox4ev3r

Solve:
\[2\tan ^{2}\theta \sin \theta - \tan ^{2}\theta=0\]
for all values in [0, 360] <---degrees

- Jamierox4ev3r

@ganeshie8 any ideas?

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## More answers

- ganeshie8

it seems \(\tan^2\theta\) is common in both the terms
you may start by factoring it out

- Jamierox4ev3r

Alright. So that would be
\( \tan( 2\tan \theta \sin \theta-\tan \theta )\)
right?

- ganeshie8

pull out the entire \(\tan^2\theta\)

- Jamierox4ev3r

the entire thing? then that would leave you with.. just sine I believe.
So
\(\tan^{2} \theta (2\sin \theta-\theta)\)

- ganeshie8

wait a sec,
\(\tan^{2} \theta (2\sin \theta-\color{Red}{\theta})\)
are you really sure that \(\color{Red}{\theta}\) stays back there ?

- Jamierox4ev3r

wait... no it wouldn't. whoops xD

- ganeshie8

\(\tan^2\theta\) is ONE SINGLE thing
\(\tan^2\) and \(\theta\) are not two pieces

- Jamierox4ev3r

^ yes I recall that now. But wouldn't something have to stay there?

- Jamierox4ev3r

oh wait. wouldn't it just be -1?

- ganeshie8

good, please fix it
also what happened to \(= 0\)

- Jamierox4ev3r

oh it's still there, I just lopped it off.
So as of now, this is where we are.
\(\tan^{2} \theta (2\sin \theta-1)=0\)

- ganeshie8

looks good!
what are you going to do next

- Jamierox4ev3r

I'm not sure....seems to me like there is a trigonometric identity somewhere in there perhaps.

- ganeshie8

Easy, actually we're almost done!
all we need to know is the "zero product property"

- ganeshie8

|dw:1441068917025:dw|

- ganeshie8

so far we have :
\(\color{red}{\tan^{2} \theta} \color{blue}{(2\sin \theta-1)}=0\)
By zero product property,
\(\color{red}{\tan^2\theta}=0\) or \(\color{blue}{2\sin\theta-1}=0\)
solve each of them separately using unit circle

- Jamierox4ev3r

so we could set either one to zero. And solve for theta \((\theta)\)

- ganeshie8

Exactly! thats what zero product property tells us

- Jamierox4ev3r

wow. that actually makes so much sense. Alright, so let me solve each separately and I'll be back with you on what I got

- ganeshie8

take ur time
you may have to use unit circle as you're asked to find all the solutions between 0 and 360

- Jamierox4ev3r

Noted.

- Jamierox4ev3r

Alright.
So I got that \(\Large\sin \theta = \frac{1}{2}\)
So from that, I found that \(\Large\theta = 30 (degrees), 150 (degrees)\)

- Jamierox4ev3r

I wasn't as sure for \(\tan \theta\), but I think that
\(\Large\tan \theta =0\)
and from that, wouldn't \(\Large\theta = 0 (degrees)\) ??
**not sure**

- ganeshie8

Excellent!
\(\tan\theta=0\) should give you 3 solutions

- ganeshie8

il give a hint :
\(\tan\theta = \tan(\theta+180)\)

- Jamierox4ev3r

oh wait i think i see. 90 degrees would be undefined

- ganeshie8

In other words :
adding 180 to the angle wont change the value of \(\tan(\theta)\)

- Jamierox4ev3r

right.

- Jamierox4ev3r

I get that.

- ganeshie8

\(\tan\theta=0\)
\(\implies \theta = \{0,\pm 180,\pm 360,\pm540,\ldots\}\)
since you want the solutions between 0 and 360
you just pick \(0,180,360\)

- Jamierox4ev3r

oh and btw, I can include 0 degrees, but not 360. I was supposed to write the problem like
[0,360) instead of [0,360]. So technically, for this problem, there are only two solutions that come from \(\tan \theta =0\)

- ganeshie8

Oh then you're absolutely right!

- Jamierox4ev3r

alright, so in total, all the solutions are 0, 30, 150, and 180! (all in degrees)
oh my goodness, thank you so much! This review material was so distant in my memory, until you came along of course. Your patience is outstanding, thank you

- ganeshie8

np
your patience is outstanding too! keep it up :)

- Jamierox4ev3r

d'aw you flatter me :')

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