SunDun111
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The Q is, solve the equation, Tan2X = 3Tanx for 0 to 2 Pi (interval),

Right I changed Tan 2X into the double angle identity, and took 3Tanx to the other side, I thought since its equal to 0 the bottom part of the fraction, of the double angle identity will cancel out but it dosent in the answer?

Btw the identity for tan2x is 2TanX/ 1- Tanx^2
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Math12345
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http://www.thestudentroom.co.uk/show...2&postcount=13
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Zacken
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(Original post by SunDun111)
The Q is, solve the equation, Tan2X = 3Tanx for 0 to 2 Pi (interval),

Right I changed Tan 2X into the double angle identity, and took 3Tanx to the other side, I thought since its equal to 0 the bottom part of the fraction, of the double angle identity will cancel out but it dosent in the answer?

Btw the identity for tan2x is 2TanX/ 1- Tanx^2
Huh? Are you talking about:

\displaystyle \tan 2x = 3\tan x \Rightarrow \frac{2\tan x}{1 - \tan^2 x} - 3\tan x = 0 \Rightarrow 2 \tan x - 3\tan x(1 - \tan^2 x) = 0

The denominator does "cancel" out when you multiply through by the denominator, but that means you need to multiply th 3 tan x by the denominator too.

The reason why \frac{a}{b} = 0 \Rightarrow \frac{a}{b} \times b = 0 \times b \Rightarrow a = 0 works is by multiplying by the denominator.

If you had \frac{a}{b} - c = 0 then \left(\frac{a}{b} - c\right) \times b = 0 \times b \Rightarrow a - bc = 0
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SunDun111
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(Original post by Zacken)
Huh? Are you talking about:

\displaystyle \tan 2x = 3\tan x \Rightarrow \frac{2\tan x}{1 - \tan^2 x} - 3\tan x = 0 \Rightarrow 2 \tan x - 3\tan x(1 - \tan^2 x) = 0

The denominator does "cancel" out when you multiply through by the denominator, but that means you need to multiply th 3 tan x by the denominator too.

The reason why \frac{a}{b} = 0 \Rightarrow \frac{a}{b} \times b = 0 \times b \Rightarrow a = 0 works is by multiplying by the denominator.

If you had \frac{a}{b} - c = 0 then \left(\frac{a}{b} - c\right) \times b = 0 \times b \Rightarrow a - bc = 0
Yeah I figured it out that was the question and i got the answers, now im kinda stuck on another,

The question is 2 Cos x^2 = 2SinxCosx + 1

How can I do this? Do i need to factorise it so i can solve two equations?
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Math12345
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Maybe this helps: 2sinxcosx+1=(sinx+cosx)^2
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Zacken
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(Original post by SunDun111)
Yeah I figured it out that was the question and i got the answers, now im kinda stuck on another,

The question is 2 Cos x^2 = 2SinxCosx + 1

How can I do this? Do i need to factorise it so i can solve two equations?
Move the one over to the other side of the equation:

2\cos^2 x - 1 = 2\sin x \cos x but this is precisely \cos 2x = \sin 2x by using your trig identities.

Hence: \tan 2x = 1 \iff 2x = \cdots
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Zacken
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(Original post by Math12345)
Maybe this helps: 2sinxcosx+1=(sinx+cosx)^2
I can't immediately see how this is helpful?
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Math12345
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(Original post by Zacken)
I can't immediately see how this is helpful?
2\cos^2(x) = (\sin(x)+\cos(x))^2
\sqrt{2}\cos(x) = \sin(x)+\cos(x) or \sqrt{2}\cos(x) = -\sin(x)-\cos(x)
\tan(x) = \sqrt{2}-1 or \tan(x) = -\sqrt{2}-1
Now solve.

Your method is better though.
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Zacken
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(Original post by Math12345)
2\cos^2(x) = (\sin(x)+\cos(x))^2
\sqrt{2}\cos(x) = \sin(x)+\cos(x) or \sqrt{2}\cos(x) = -\sin(x)-\cos(x)
\tan(x) = \sqrt{2}-1 or \tan(x) = -\sqrt{2}-1
Now solve.

Your method is better though.
I see, bit long winded but it'd do. Thanks.
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