if 0 degrees < x<360 degrees, solve the equation sin x= -sqrt 3/2
what?
that makes no sense?
sorry made a mistake 4pi/3
but yea use the unit circle
sinx=sin(-π/3) ,(0,2π) x=2kπ - (π/3) or x=2kπ + π -(π/3) x=2kπ + (2π/3) k∈Z 0<x<2π 0<2kπ - (π/3)<2π or 0<2kπ + (2π/3)<2π solve the above inequalities for k (k is an integer)
so the answer is 150 degrees and 300 degrees?
\(\bf sin(x)= -\sqrt{\frac{3}{2}}\quad \textit{taking }sin^{-1}\textit{ to both sides}\\ \quad \\ sin^{-1}[sin(x)]= sin^{-1}\left(-\sqrt{\frac{3}{2}}\right)\implies x= sin^{-1}\left(-\sqrt{\frac{3}{2}}\right)\)
\(\bf \textit{recall that }sin^{-1}[sin(\theta)]=\theta\)
\(\bf sin^{-1}\left(-\sqrt{\frac{3}{2}}\right)\implies \textit{the angle(s) whose sine is }-\sqrt{\frac{3}{2}}\)
so check your unit circle
okay so then its 120 and 240? in degrees
240 is in the 3rd quadrant, sine is negative there, yes 120 is in the 2nd quadrant, sine is positive there, no dice
notice, is a negative \(\Large -\sqrt{\frac{3}{2}}\)
soo then it has to be 300 then
yeap
thank you
lemme recheck something
ok
ohh... I see.. nevermind... you meant ... to write I gather \(\bf sin(x)= -\cfrac{\sqrt{3}}{2}\quad \textit{taking }sin^{-1}\textit{ to both sides}\\ \quad \\ sin^{-1}[sin(x)]= sin^{-1}\left(-\cfrac{\sqrt{3}}{2}\right)\implies x= sin^{-1}\left(-\cfrac{\sqrt{3}}{2}\right)\\ \quad \\ x= \begin{cases} 240^o\\ \bf 300^o \end{cases}\) just making the distinction, that \(\bf -\cfrac{\sqrt{3}}{2}\ne -\sqrt{\cfrac{3}{2}}\)
actually the signs "<" are suppose to be greater that or equal to for both so wold it still be 240 and 300?
0 < x < 360 just means 0 < x <--- "x" is greater than 0 x < 360 <---- "x" is lesser than 360 well, 240 is greater than 0 and less than 360 300 is also greater than 0 and less than 360
they're not BELOW 0 or equals it, they're not ABOVE 360 or equals it
240 degrees
ok so then it is 240 and 300
yes
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