Find all solutions to the equation::

(sin x- cos x)^2=3

Question

Find all solutions to the equation::

(sin x- cos x)^2=3

reemii · Answer

your first exercise told you that (sin x - cos x)^2 = 1 - sin(2x), so use that and transform the equation.

anonymous · Answer

So would we put 1-sin(2x) equal to 3? Because you can cross out the first part of the equation with the part you just examples

reemii · Answer

"So would we put 1-sin(2x) equal to 3"  <- exactly, so after that you can use usual transformations to isolate "sin(2x)".

anonymous · Answer

So then you would have -sin(2x)=4 and then times by a negaitve giving you sin(2x)=-4. Then you can plug it into your calculator and figure out the answers?

reemii · Answer

in these type of equations, when only one trigonometric function remains, you usually want to isolate it, because something that looks like "sin(stuff) = 1" tells you something about "stuff".

anonymous · Answer

We should take the inverse of both sides and then 2x= sin^-1(4) right?

anonymous · Answer

Then you would divide by 2 and then i'm lost....

reemii · Answer

wait, 1-sin(2x) = 3 <=> -sin(2x) = 2 <=> sin(2x) = -2. Here, it looks like there's nothing to compute anymore: what you got is "sin(something) = -2", that is impossible.

anonymous · Answer

Ok, so then we will need to leave it that way.... A sin product can be negative in this case?

anonymous · Answer

actually it can be negative; but the thing is that sin value cannot exceed 1 and hence in any other quadrant, can also not be less than -1. so no solution.

reemii · Answer

what is "sin product" ?
hmm, what I see is that "sin(whatever)" can never be equal to -2, it always some number between -1 and 1. Therefore you say: "no solution".

reemii · Answer

he just said it lol.

anonymous · Answer

Yea, great minds think alike. So for the next question it asks: find all solutions for (sin x- cos x)^2=1.....

anonymous · Answer

this one is quite simple.

you can expand again and solve; come till where you're left only with 2 sin x cos x.

anonymous · Answer

That one wouldn't work either then correct? because -sin(2x) can't equal 0... because I put 1-sin(2x)=1 again like before and then subtraacted one

anonymous · Answer

LOL of course sin 2x can equal 0.

bcoz 0 is sin 0

so sin 2x = sin 0

2x = 0.

anonymous · Answer

Ooook, but you just said that sin has to either equal -1 or 1 to be a solution... So 0 can work as well for these problems?>

anonymous · Answer

@HemerickLee  he said BETWEEN -1 and 1.

anonymous · Answer

Ahhh, i'm sorry wrong reading.... Makes since now... So they can only equal -1,0,1 to be a solution available

reemii · Answer

when seeing "sin(2x)=0", all you need to thing of first is "can sin(soemthing) be equal to 0"?
If the answer is yes (it is), then you conclude something about "something".
"sin(something) = 0" means "something=0 or pi or 2pi ... or -pi or -2pi or ..".

Correctly: $\sin (2x) = 0 \iff 2x = k\pi,$ for $k\in\mathbb Z (\text{integers}) \iff x = \frac{k\pi}{2}$ with $k$ integer.

anonymous · Answer

@reemii  so isn't it simply 0. ?

anonymous · Answer

So x= $$\frac{ \pi n }{ 2 }$$

reemii · Answer

there's more than just 0 as solution, since sin(0) = sin(2pi) = sin(4pi) etc..

anonymous · Answer

Like my answer says... It can be any thing with pi after it?

reemii · Answer

"x = n pi / 2" means, every x of the form $n\pi/2$ (with $n$ integer) is a solution.

reemii · Answer

what do you mean by "any thing with pi after it" ? (sounds dangerous..)

anonymous · Answer

Alright, Thank you guys! psssswwwwhhh glad to get that off of my chest...