Math, asked by g6m12, 1 year ago

Find the general solution of tan2Ѳ + ( 1- root3 ) tanѲ - root3 = 0

KARG1L: Hope it was to find tan⊙=

AvmnuSng: is it tan(2A) or tan^2(A), means tan_square_theta or tan_2_theta , what's the first term, i guess it's tan_2_theta

g6m12: its tan square theta...

AvmnuSng: for next tym, write any squared term as tan^2(a)... like this

KARG1L: I answered it right then

g6m12: k...

Answers

Answered by KARG1L

See image.....I solved on paper.

Attachments:

g6m12: thnx....

AvmnuSng: general solution dude?? where's the general soln..

KARG1L: I found tan thita

AvmnuSng: u hv to find theta, general soln

KARG1L: M done with first term of c

KARG1L: Lass 10 and I still don't understand the term. General solution.

KARG1L: I did not find it nywhere

KARG1L: Sorry

Answered by AvmnuSng

$tan^2(\theta) + (1 - \sqrt{3}) tan(\theta) - \sqrt{3} = 0 \\ tan^2(\theta) + tan(\theta) - \sqrt{3} tan(\theta) - \sqrt{3} = 0 \\ tan(\theta)(tan(\theta) + 1) - \sqrt{3}(tan(\theta) + 1) = 0 \\ (tan(\theta) + 1)(tan(\theta) - \sqrt{3} ) = 0 \\ tan(\theta) = (-1, \sqrt{3}) \\ \\ \theta = (\frac{3 \pi }{4}, \frac{ \pi }{3})$

So general Solution
$\theta = (n \pi + \frac{3 \pi }{4}, n \pi - \frac{3 \pi }{4}, n \pi + \frac{\pi }{3}, n \pi - \frac{\pi }{3})$

KARG1L: Hey, can u tell me what is general solution??

AvmnuSng: the value of sin of (30, 390, 750, 1110, ...) all is equal to 1/2. so if i say solve sin(a) = 1/2, then a = 30 is one solution, but all the other solutions can't be written coz they are countably infinite, but you can say a = n * 360 + 30, and n = 0, 1, 2, 3, ....... Now this is general solution ....

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