Question

Find all solutions to the equation in the interval [0, 2π).

cos x = sin 2x

Answer 1

is this
\[\cos(x)=\sin(2x)\] or is it
\[\cos(x)=\sin^2(x)\]?

Answer 2

The first one!

Answer 3

use the identity
\[\sin(2x)=2\sin(x)\cos(x)\] to rewrite it as
\[\cos(x)=2\sin(x)\cos(x)\]

Answer 4

or better yet
\[2\sin(x)\cos(x)-\cos(x)=0\] or still better
\[\cos(x)\left(2\sin(x)-1\right)=0\]then it will be much easier

Answer 5

How on Earth did you do that? How do cos(x) and sin(2x) equal the same thing?

Answer 6

step one is an identity \[\sin(2x)=2\sin(x)\cos(x)\] always

Answer 7

step to is taking
\[\cos(x)=2\sin(x)\cos(x)\] and set it equal to zero

Answer 8

Wouldn't it go to 2sinx=0?

Answer 9

subtract \(\cos(x)\) from both sides and get
\[2\sin(x)\cos(x)-\cos(x)=0\]

Answer 10

Oh, my bad. Confusion between the multiplication and addition. What then?

Answer 11

How do you find solutions for one like this? Factor?