Asked by Anonymous
sin2x+cosx=0 , [-180,180)
= 2sinxcosx+cosx=0
= cosx(2sinx+1)=0
cosx=0
x1=cos^-1(0)
x1=90
x2=360-90
x2=270
270 doesn't fit in [-180,180) what do I do? Or maybe I did something wrong.
sinx=1/2
x1=sin^-1(1/2)
x1=30
x2=180-30
x2=150
is this correct?
Please and Thank you
= 2sinxcosx+cosx=0
= cosx(2sinx+1)=0
cosx=0
x1=cos^-1(0)
x1=90
x2=360-90
x2=270
270 doesn't fit in [-180,180) what do I do? Or maybe I did something wrong.
sinx=1/2
x1=sin^-1(1/2)
x1=30
x2=180-30
x2=150
is this correct?
Please and Thank you
Answers
Answered by
Reiny
but your domain is from -180 to 180 , so even though 270 will work in the equation , it is beyond your domain.
But, by looking at the cosine graph , you will see that it also has an x-intercept at x = -90°
so for your first one,
x = -90, 0, 90
the second part for sinx = 1/2
your answers of 30° or 150° are correct for the given domain.
But, by looking at the cosine graph , you will see that it also has an x-intercept at x = -90°
so for your first one,
x = -90, 0, 90
the second part for sinx = 1/2
your answers of 30° or 150° are correct for the given domain.
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