Question

sin x - sqrt3 cos x=1

Answer 1

write as a single function of sine

Answer 2

thank you but that is where I am getting confused. lol

Answer 3

\[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\] where
\[\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}\] and
\[\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\]

Answer 4

sometimes you can use
\[\theta=\tan^{-1}(\frac{b}{a})\] but not always

Answer 5

you good from there? this one is cooked up to be more or less easy

Answer 6

not really these problems I tend to make ten times harder than it needs to be.

Answer 7

in your case \(a=1,b=\sqrt3\) so \(\sqrt{a^2+b^2}=\sqrt{1+3}=\sqrt4=2\)

Answer 8

actually \(b=-\sqrt3\) but it makes no difference when you square it

Answer 9

as for \(\theta\), you have to find a number (angle if you prefer) where
\[\sin(\theta)=\frac{b}{a}=-\frac{\sqrt3}{2}\] and
\[\cos(\theta)=\frac{1}{2}\] this should look familiar

Answer 10

oops i did not mean \(\sin(\theta)=\frac{b}{a}\) i meant \(\sin(\theta)=\frac{ b}{\sqrt{a^2+b^2}}\)

Answer 11

so if the answer needs to be on an interval from 0 to 2pi i would add the period to the answer until i reach 2pi?

Answer 12

we are not close to done yet
what is \(\theta\)?

Answer 13

oh lol they dont have give what theta is

Answer 14

the instruction say give the answer in exact form on the interval of [0, 2pi)
the question itself is sin theta - square root 3 cos theta = 1

Answer 15

you can square both sides to get \(sin^2 -2\sqrt{3} sin~ cos +3 cos^2 =1\) and separate to \(sin^2 + cos^2 +2 cos^2 -2 \sqrt{3}sin~ cos=1\), you have \(sin^2 + cos^2 =1\) cancel out with the right handside, therefore, just \(2cos^2 -2\sqrt{3}sin~cos=0\)
factor 2cos out, \(2cos (cos -\sqrt{3}sin) =0\)
so, cos =0 or \(cos -\sqrt{3}sin =0\)
in that interval, cos =0 iff \(\theta =\dfrac{\pi}{2}~ or ~\theta=\dfrac{3\pi}{2}\)
\(cos - \sqrt{3}sin =0\\cos = \sqrt{3}sin\\\dfrac{cos}{sin}= \sqrt{3}\\cot =\sqrt{3}\)
you do the leftover

Answer 16

thank you very much for your help