Question

sqrt(3)sinx+cosx=1 Please help!! :) Thanks

Answer 1

write as a single function of sine first

Answer 2

\[a\sin(x)+b\cos(x)=r\sin(x+\theta)\] where \(r=\sqrt{a^2+b^2}\) and \(\tan(\theta)=\frac{b}{a}\)
does this look familiar? we can walk through it if you like

Answer 3

Yeah, that looks familiar. Do you take the cosx to the other side??

Answer 4

oh no

Answer 5

we are going to get \[\sqrt3\sin(x)+\cos(x)=r\sin(x+\theta)=1\] and solve that
we need to find \(r\) and \(\theta\) first

Answer 6

this one has been cooked up to be not too hard
\[r=\sqrt{\sqrt{3}^2+1^2}=\sqrt4=2\]

Answer 7

and since \(\sin(\theta)=\frac{b}{r}=\frac{1}{2}\) and \(\cos(\theta)=\frac{a}{r}=\frac{\sqrt3}{2}\) you get \(\theta=\frac{\pi}{6}\)

Answer 8

now your job is to solve
\[2\sin(x+\frac{\pi}{6})=1\]

Answer 9

I got the answer to be 2pi/3 + (2pi)k and 0+(2pi)k

Answer 10

looks good to me

Answer 11

Thanks!! :)