Question

Find all solutions to the equation in the interval [0, 2π).
cos x = sin 2x

Answer 1

sin2x=2sinxcosx
so:
cosx=2sinxcosx
or
1=2sinx
or
sinx=1/2
this is true for x=?

Answer 2

\(\bf cos(x)=sin(2x)\implies cos(x)=2sin(x)cos(x)
\\ \quad \\
\cfrac{\cancel{cos(x)}}{\cancel{cos(x)}}=2sin(x)\implies \cfrac{1}{2}=sin(x)
\\ \quad \\
sin^{-1}\left( \cfrac{1}{2} \right)=sin^{-1}[sin(x)]\implies ?\)

Answer 3

i got 0, pi/6, 5pi/6, pi

Answer 4

hmmm

Answer 5

there is a mistake in the above work
\[\cos x=2\sin x\cos x \Longrightarrow \cos x-2\sin x \cos x=0\]
\[\cos x(1-2\sin x)=0 \Longrightarrow \cos x=0 ~~or~~\sin x= 1/2\]

Answer 6

you're correct actually
the 0 is part of the angles

Answer 7

right.... should have taken common factor
and that'd give a 0 value for the angles

Answer 8

cancellation with cos kills some solutions because it assumes cosx is not zero

Answer 9

i got 0, pi/6, 5pi/6, pi \(\huge \checkmark\)

Answer 10

should be good:)

Answer 11

thanks guys :)

Answer 12

yw

Answer 13

np