Can you solve each of those two equations now?

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- Aug 17th 2011, 06:25 PM #1

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## (tan x)^2 = sin 2x

anyone has an idea on how to solve for this equation, (tan x)^2 = sin 2x

I know that sin 2x = 2sin xcos x and tan x = sin x/cos x, how ever i am having trouble simplifyng the equation into a single function so that I can solve for x.

thanks in advance

@sammys thanks for your advice on the other thread.

- Aug 17th 2011, 06:48 PM #2

- Aug 17th 2011, 09:57 PM #3

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- Aug 18th 2011, 02:23 AM #4

- Aug 18th 2011, 04:51 PM #5

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## Re: (tan x)^2 = sin 2x

@prove it..

I arrived at the fourth line of the solution you had presented, and i never thought of factoring out sin x, i missed that part. yes, I can solve the remaining equations now. a million thanks to you.

to otherw whoe replied, thank you.

- Aug 18th 2011, 04:55 PM #6

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- Aug 18th 2011, 05:53 PM #7

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- Aug 18th 2011, 06:18 PM #8

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## Re: (tan x)^2 = sin 2x

Divide by cos(x), which is fine if cos(x) ≠ 0. And clearly, cos(x) = 0 is not a solution to this equation.

sin(x)/cos(x) = tan(x).

Multiply by sec^2(x), and change sec^2(x) to 1 + tan^2(x).

You then have a cubic equation in tan(x). Let u = tan(x), & factor to solve for u.

- Aug 19th 2011, 09:19 AM #9

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- Aug 19th 2011, 09:35 AM #10