Question

find all values of x in the interval[0,2pie] that satisfies the equation sin2x=cosx

Answer 1

pi/2 and 3pi/2

Answer 2

We have
\[\sin 2x= \cos x\]
\[2 \sin x \cos x-\cos x=0\]
so
\[\cos x(2 \sin x -1)=0\]
either cos x=0 or 2 sin x-1=0=> sin x =1/2
cos x=0
\[x= \pi/2, 3\pi/2\]
sin x=1/2
\[x=\pi/6, (\pi-\pi /6)=>\pi/6, 5\pi /6\]

so \[x= \pi/6, \pi/2, 5\pi/6, 3\pi/2\]

Answer 3

this is because the interval is [0,2pi] right

Answer 4

yeah, otherwise there'd be many solutions

Answer 5

ok again thank you so much :)

Answer 6

oops, I read it as sin(2x)=cos(x)=0, missed the two other sol'ns. good job getting the other two.