Question

Solve:

Answer 1

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

Answer 2

Solve:
\[2\tan ^{2}\theta \sin \theta - \tan ^{2}\theta=0\]

for all values in [0, 360] <---degrees

Answer 3

@ganeshie8 any ideas?

Answer 4

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

Answer 5

Alright. So that would be

\( \tan( 2\tan \theta \sin \theta-\tan \theta )\)

right?

Answer 6

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

Answer 7

the entire thing? then that would leave you with.. just sine I believe.

So

\(\tan^{2} \theta (2\sin \theta-\theta)\)

Answer 8

wait a sec,


\(\tan^{2} \theta (2\sin \theta-\color{Red}{\theta})\)

are you really sure that \(\color{Red}{\theta}\) stays back there ?

Answer 9

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

Answer 10

\(\tan^2\theta\) is ONE SINGLE thing

\(\tan^2\) and \(\theta\) are not two pieces

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

oh it's still there, I just lopped it off.

So as of now, this is where we are.

\(\tan^{2} \theta (2\sin \theta-1)=0\)

Answer 15

looks good!
what are you going to do next

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Exactly! thats what zero product property tells us

Answer 21

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

Answer 22

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

Answer 23

Noted.

Answer 24

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

Answer 25

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**

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

right.

Answer 31

I get that.

Answer 32

\(\tan\theta=0\)
\(\implies \theta = \{0,\pm 180,\pm 360,\pm540,\ldots\}\)

since you want the solutions between 0 and 360
you just pick \(0,180,360\)

Answer 33

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 \(\tan \theta =0\)

Answer 34

Oh then you're absolutely right!

Answer 35

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

Answer 36

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

Answer 37

d'aw you flatter me :')