Question

Solve for the indicated domain:

cosx + sqrt3 sinx=1 {0, 2pi}

Answer 1

write as a single function first

Answer 2

\[\cos x+\sqrt{3} \sin x=1 ; x \in [0, 2\Pi]\]

Answer 3

first write
\[\cos(x)+\sqrt3\sin(x)\] as a single sine function

Answer 4

do you know what i mean here?

Answer 5

Not really.......

Answer 6

ok
you can always turn
\[b\cos(x)+a\sin(x)\] in to a single sine function

Answer 7

we can walk through it if you like
you are going to take
\[\cos(x)+\sqrt{3}\sin(x)=k\sin(x+\theta)\]
what you need is \(k\) and \(\theta\)

Answer 8

it is a direct consequence of the addition angle formula for sine, but you don't need the derivation, just the mechanics

Answer 9

ready?

Answer 10

Yep

Answer 11

\[k=\sqrt{a^2+b^2}=\sqrt{1^2+\sqrt{3}^2}=\sqrt{1+3}=\sqrt{4}=2\]

Answer 12

I got that!
Yay

Answer 13

k now for \(\theta\)
use
\[\sin(\theta)=\frac{b}{k}\] and here it is important to remember that \(b\) is the coefficient of the COSINE term, so we know
\[\sin(\theta)=\frac{1}{2}\]

Answer 14

also
\[\cos(\theta)=\frac{a}{k}=\frac{\sqrt3}{2}\]

Answer 15

so what it \(\theta\) ?

Answer 16

.............That's where i got stuck

Answer 17

you know an angle where the sine is one half and the cosine is root three over two??

Answer 18

\[\Pi/6\] or 30 degrees

Answer 19

yeah
lets be adults and stick with number rather than degrees
so we get
\[\cos(x)+\sqrt3\sin(x)=2\sin(x+\frac{\pi}{6})\]

Answer 20

you can check this is right by using the addition angle formula for the right hand side and see that you get the left hand side
if you like
now your job is to solve
\[2\sin(x+\frac{\pi}{6})=1\] which should take only two or three steps

Answer 21

you good from there? or still need help?

Answer 22

oh i was doing something. Back now though @~@

Answer 23

lol
pretty sophisticated problem for high school math
especially in the summer

Answer 24

Yep.

Answer 25

Question: Do you have to divide both sides by 2?

Answer 26

yes

Answer 27

So then I get\[\sin x \cos \Pi/6+\sin \Pi/6\cos x\]

Answer 28

you lost me there

Answer 29

\[2\sin(x+\frac{\pi}{6})=1\] divide by 2 and get
\[\sin(x+\frac{\pi}{6})=\frac{1}{2}\]

Answer 30

I did, then I expanded the left side

Answer 31

don't do that
that is working backwards

Answer 32

if
\[\sin(x+\frac{\pi}{6})=\frac{1}{2}\] then
\[x+\frac{\pi}{6}=\frac{\pi}{6}\] and so \(x=0\) one solution

Answer 33

Oh. Derp

Answer 34

then there is probably another
for example \(\sin(\frac{5\pi}{6})=\frac{1}{2}\) and so you can solve
\[x+\frac{\pi}{6}=\frac{5\pi}{6}\] for \(x)\)

Answer 35

x= 2pi/ 3 right

Answer 36

yeah i think that looks good

Answer 37

Ok thank you so much! I have a test today and I needed help OMG THANK YOU

Answer 38

yw
good luck!