Question

find the exact solutions to the equation in the interval [0,2pi): -3sin4x= -3cos2x

Answer 1

@satellite73 please help me :(

Answer 2

I would consider transforming the 4x into an equivalent 2x.

sin(2x) = 2sin(x)cos(x) so sin(4x) = 2sin(2x)cos(2x)

Answer 3

but how will i go from there i m confused i did this problem and i got to the part where i am sin4x-1-2sin^2x=0 can u help am i doing anything wrong?

Answer 4

You did not do as I suggested. It will look MUCH better if you do. Leave the cosine side alone.

-3sin(4x) = -3cos(2x)

What's up with those -3s?

sin(4x) = cos(2x)

Now, apply the double angle formula

2sin(2x)cos(2x) = cos(2x)

Now what?

Answer 5

I guess I should clarify the original problem statement.

Is the right side \(\cos(2x)\) or \(\cos^{2}(x)\)?

Answer 6

cos(2x)

Answer 7

Perfect. Then, we're on the right track. Do you see how to proceed?

\(2\sin(2x)\cos(2x) = \cos(2x)\)

Answer 8

and i tried simplifying that and i got sin2x=1/2? because i divided co(2x) both sides and it cancels out to 1 then i divide by 2 so i get sin2x=1/2?

Answer 9

No good. Never do that. FACTOR ONLY. You missed solutions.

2sin(2x)cos(2x) - cos(2x) = 0

(2sin(2x) - 1)cos(2x) = 0

sin(2x) = 1/2 OR cos(2x) = 0

Answer 10

oh i get it!! thank you! i can solve it by factoring and so the answer i get would be pi/6 5pi/6 and pi/2 and 3pi/2

Answer 11

then i multiply those by 2!

Answer 12

THERE they are. Those missing ones were lonely until you rescued them. Good work.

\(\cos(2x) = 0 \implies 2x = \dfrac{\pi}{2} + k\pi\) where k is an integer.

Finally, \(x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\). Just pick all the integers where this value is in the desired Domain.

Do the same with \(\sin(2x) = 1/2\) and you have them all.

Answer 13

thank you!!!!! you were soo helpful! i have one more if u dont mind helping me understand that one its cot and csc

Answer 14

Post and tag. It's past my bedtime, so no dilly-dallying. :-)

Answer 15

well thanks anyways for the help!! :D