Solving equations

**arze** · July 3rd 2009, 01:03 AM

Solve the following equation giving values for 0-360deg
cosA=cos2A+cos4A

cos2A=cos(30-A)

i probably missed something on the identities part.
thanks

**alexmahone** · July 3rd 2009, 01:21 AM

$cos A=cos 2A+cos 4A$

$cos A=2cos 3A cos A$

$cos A=0$ or $cos 3A=\frac{1}{2}$

$A=90^0$ or $3A=60^0$

$A=90^0$ or $A=20^0$

**running-gag** · July 3rd 2009, 01:23 AM

Originally Posted by arze

Solve the following equation giving values for 0-360deg
cosA=cos2A+cos4A

Hi

You can use the trig identity $\cos p + \cos q = 2 \cos \left(\frac{p+q}{2}\right) \: \cos \left(\frac{p-q}{2}\right)$

**arze** · July 3rd 2009, 01:27 AM

ah! that's what i missed in the first one. thanks!
how about the second one?

**alexmahone** · July 3rd 2009, 01:34 AM

Originally Posted by arze

ah! that's what i missed in the first one. thanks!
how about the second one?

$cos 2A = cos (30-A)$

$2A=30^o-A$

$3A=30^o$

$A=10^o$

OR

$2A=A-30$

$A=-30^0$

**arze** · July 3rd 2009, 01:39 AM

Originally Posted by alexmahone

$cos 2A = cos (30-A)$

$2A=30^o-A$

$3A=30^o$

$A=10^o$

OR

$2A=A-30$

$A=-30^0$

i'm supposed to give the answers $10^o$ , $130^o$ , $250^o$ , and $330<br /> ^o$
i can understand $10^o$ and $330^o$ but what about the other two?

**yeongil** · July 3rd 2009, 02:42 AM

Originally Posted by arze

i'm supposed to give the answers $10^o$ , $130^o$ , $250^o$ , and $330<br /> ^o$
i can understand $10^o$ and $330^o$ but what about the other two?

Looking back at alexmahone's work:

Originally Posted by alexmahone

$cos 2A = cos (30-A)$

$2A=30^o-A$

$3A=30^o$

If A is supposed to be between 0° and 360°, then 3A should be between 0° and 1080° (360° times 3). So from here you also have to mention that

$3A=390^{\circ}$ and

$3A=750^{\circ}$

because these angles are coterminal to 30° and they are between 0° and 1080°. Now divide both sides of both equations by 3 and you will get

$A=130^{\circ}$ and

$A=250^{\circ}$ .

01

**arze** · July 3rd 2009, 02:59 AM

how could i have missed that?
thank you very much!

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