Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

find the values of x in the interval [x,2pi] that satisfy sin^2x + sinx - 2 = 0

OpenStudy (anonymous):

(sin(x) +2)(sin(x)-1)

OpenStudy (anonymous):

how did you get that

OpenStudy (anonymous):

(sin(x) +2)(sin(x)-1)=0

OpenStudy (anonymous):

Just factor it. How do you factor \[u^2 +u-2=(u+2)(u-1)\]

OpenStudy (anonymous):

Since sin(x) cannot equal 2 then sin(x) =1 hence \[ x=\frac \pi 2 + 2\ k \pi \]

OpenStudy (anonymous):

Did you get it?

OpenStudy (anonymous):

im lost at sinx not = to 2

OpenStudy (anonymous):

Sin x varies between 0 and 1 Solutions in interval [x,2pi] required by the question

OpenStudy (anonymous):

So \[ x=\frac \pi 2 \]

OpenStudy (anonymous):

what does k stand for?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!