# Thread:

1. ## Solving equations

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

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

$cos A=2cos 3A cos A$

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

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

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

3. Originally Posted by arze
Solve the following equation giving values for 0-360deg
cosA=cos2A+cos4A
Hi

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

4. ah! that's what i missed in the first one. thanks!

5. Originally Posted by arze
ah! that's what i missed in the first one. thanks!
$cos 2A = cos (30-A)$

$2A=30^o-A$

$3A=30^o$

$A=10^o$

OR

$2A=A-30$

$A=-30^0$

6. 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?

7. 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

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