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

$\displaystyle cos A=2cos 3A cos A$

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

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

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

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

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

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!
$\displaystyle cos 2A = cos (30-A)$

$\displaystyle 2A=30^o-A$

$\displaystyle 3A=30^o$

$\displaystyle A=10^o$

OR

$\displaystyle 2A=A-30$

$\displaystyle A=-30^0$

6. Originally Posted by alexmahone
$\displaystyle cos 2A = cos (30-A)$

$\displaystyle 2A=30^o-A$

$\displaystyle 3A=30^o$

$\displaystyle A=10^o$

OR

$\displaystyle 2A=A-30$

$\displaystyle A=-30^0$
i'm supposed to give the answers $\displaystyle 10^o$, $\displaystyle 130^o$, $\displaystyle 250^o$, and $\displaystyle 330 ^o$
i can understand $\displaystyle 10^o$ and $\displaystyle 330^o$ but what about the other two?

7. Originally Posted by arze
i'm supposed to give the answers $\displaystyle 10^o$, $\displaystyle 130^o$, $\displaystyle 250^o$, and $\displaystyle 330 ^o$
i can understand $\displaystyle 10^o$ and $\displaystyle 330^o$ but what about the other two?
Looking back at alexmahone's work:
Originally Posted by alexmahone
$\displaystyle cos 2A = cos (30-A)$

$\displaystyle 2A=30^o-A$

$\displaystyle 3A=30^o$
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

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

$\displaystyle 3A=750^{\circ}$

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

$\displaystyle A=130^{\circ}$ and

$\displaystyle A=250^{\circ}$.

01

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