Question

Solve for X, which is a real number.

sin 2x = sinx , 0<= x <= 2pi


lny = 2t - 3

Answer 1

you should have a separate post for each question

Answer 2

Ehh,.. same topic, same post I feel.

Answer 3

Observe that \(\sin(2x)=\sin(x)\) is only true whenever \(x=\pi k\), for all integers \(k\).

Answer 4

My mistake, that is only a subset of the solutions. What is the other subset?

Answer 5

\[ 0 \le x \le 2\pi \]

Answer 6

That's the restriction.

We already found that \(x=\pi k\), \(k\in\mathbb{Z}\) is a part of the solution set. The other parts would be \(x=2\pi k+\pi/3\) and \(x=2\pi k-\pi/3\), for \(k\in\mathbb{Z}\).

Answer 7

Hence, what are your three solutions in the interval \(0\leqslant x\leqslant2\pi\)?

Answer 8

x = 0 or pi
x = pi/3
x = pi/3 or 5/3pi

?

Answer 9

You can write sin(2x) as 2sin(x)cos(x) and then move that little other sin(x) fella over with it's other trig buddies and do some factoring.
Recall if ab=0 then either a=0 or b=0 or both=0.

Answer 10

That's what I just did and I got the answers above, I believe this is correct ,so we're done here.