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

cos x = sin 2x

pi divided by two., three pi divided by two.
pi divided by six., pi divided by two., five pi divided by six., three pi divided by two.
0, π
0, pi divided by six, five pi divided by six., π

Question

Find all solutions to the equation in the interval [0, 2π).

cos x = sin 2x

 pi divided by two., three pi divided by two.
 pi divided by six., pi divided by two., five pi divided by six., three pi divided by two.
 0, π
 0, pi divided by six, five pi divided by six., π

laughoutloud · Answer

@mathmate

laughoutloud · Answer

@pooja195

phi · Answer

do you know the identity for sin(2x) ?

laughoutloud · Answer

no

zepdrix · Answer

Hello miss :)
Welcome to OpenStudy!

So our Sine Double Angle Formula looks like this:
$\large\rm sin2x=2sin x cos x$

zepdrix · Answer

So we have,
$\large\rm cos x=sin2x$
$\large\rm cos x=2sin x cos x$

zepdrix · Answer

Generally speaking, when trying to find solutions,
`subtraction` is better than `division`.

You might notice that both sides of the equation have a cosx in them.
So why not divide both sides by cosx?
Well because that leads to problems.
We would actually be dividing out some of our solutions for x.

Instead let's subtract cos x from each side.
$\large\rm 0=2sin x cos x-cos x$

zepdrix · Answer

From there, we can do some factoring,
and then apply our Zero Factor Property.

What do you think? Confusion? :)