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trig questions c4

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
http://www.thestudentroom.co.uk/showpost.php?p=1649342&postcount=13
Reply 2
Avatar for Zacken
Zacken
22
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:

tan⁡2x=3tan⁡x⇒2tan⁡x1−tan⁡2x−3tan⁡x=0⇒2tan⁡x−3tan⁡x(1−tan⁡2x)=0\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 ab=0⇒ab×b=0×b⇒a=0\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 ab−c=0\frac{a}{b} - c = 0 then (ab−c)×b=0×b⇒a−bc=0\left(\frac{a}{b} - c\right) \times b = 0 \times b \Rightarrow a - bc = 0
Reply 3
Avatar for SunDun111
SunDun111
OP
11
Original post by Zacken
Huh? Are you talking about:

tan⁡2x=3tan⁡x⇒2tan⁡x1−tan⁡2x−3tan⁡x=0⇒2tan⁡x−3tan⁡x(1−tan⁡2x)=0\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 ab=0⇒ab×b=0×b⇒a=0\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 ab−c=0\frac{a}{b} - c = 0 then (ab−c)×b=0×b⇒a−bc=0\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?
Maybe this helps: 2sinxcosx+1=(sinx+cosx)^2
Reply 5
Avatar for Zacken
Zacken
22
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:

2cos⁡2x−1=2sin⁡xcos⁡x2\cos^2 x - 1 = 2\sin x \cos x but this is precisely cos⁡2x=sin⁡2x\cos 2x = \sin 2x by using your trig identities.

Hence: tan⁡2x=1  ⟺  2x=⋯\tan 2x = 1 \iff 2x = \cdots
Reply 6
Avatar for Zacken
Zacken
22
Original post by Math12345
Maybe this helps: 2sinxcosx+1=(sinx+cosx)^2


I can't immediately see how this is helpful?
Original post by Zacken
I can't immediately see how this is helpful?


2cos⁡2(x)=(sin⁡(x)+cos⁡(x))22\cos^2(x) = (\sin(x)+\cos(x))^2
2cos⁡(x)=sin⁡(x)+cos⁡(x)\sqrt{2}\cos(x) = \sin(x)+\cos(x) or 2cos⁡(x)=−sin⁡(x)−cos⁡(x)\sqrt{2}\cos(x) = -\sin(x)-\cos(x)
tan⁡(x)=2−1\tan(x) = \sqrt{2}-1 or tan⁡(x)=−2−1\tan(x) = -\sqrt{2}-1
Now solve.

Your method is better though.
Reply 8
Avatar for Zacken
Zacken
22
Original post by Math12345
2cos⁡2(x)=(sin⁡(x)+cos⁡(x))22\cos^2(x) = (\sin(x)+\cos(x))^2
2cos⁡(x)=sin⁡(x)+cos⁡(x)\sqrt{2}\cos(x) = \sin(x)+\cos(x) or 2cos⁡(x)=−sin⁡(x)−cos⁡(x)\sqrt{2}\cos(x) = -\sin(x)-\cos(x)
tan⁡(x)=2−1\tan(x) = \sqrt{2}-1 or tan⁡(x)=−2−1\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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