Question

please help! Find all solutions to the equation (sin x–cos x)²=3

Answer 1

\[(a-b)^2=a^2-2ab+b^2\]Apply this.
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Answer 2

can you show me?

Answer 3

You can let a=sin(x) and b=cos(x), try. :) then, substitute after. tell me, what did you get.

Answer 4

\[\large (\sin x -\cos x)^2-3=0\]
\[\large \sin^2 x-2\sin x \cos x+\cos^2 x-3=0\]
\[\large \sin^2 x-2\sin x\cos x+1-\sin^2 x-3=0\]
\[\large -2\sin x\cos x-2=0\]
\[\large 2\sin x\cos x=-2\]
Using your trig identities:
\[\large \sin 2x=2\sin x\cos x\]
Therefore:
\[\large \sin 2x=-2\]
\[\large 2x= no solution\]

Answer 5

We really should spend more time on this transformation. It is quite useful...

\(\sin(x) - \cos(x) = \sqrt{2}\cdot \cos\left(\dfrac{3\pi}{4} - x\right)\)

Answer 6

That's where I was gonna follow up to say use auxillary angles.

Answer 7

ANd you got it wrong way around inside the cosine.

Answer 8

thank you all so much, i will go over this now. i dont know who to give best answer wish i could give it to you both

Answer 9

From @Azteck \[\sin(2x)=-2\]\[2x=\sin^{-1}(-2)\]\[x=\frac{ \sin^{-1} (-2) }{ 2 }\]
Inverse function of sine is \[-1\le x \le1\]So, \[\sin^{-1} (-2)=undefined\]OR it's not in the domain

Answer 10

\[\sin x -\cos x=\sqrt{2}\sin(\frac{x-\pi}{4}=\sqrt{3}\]

Answer 11

Wth... computer error. Sorry wrote it wrong.

Answer 12

@Azteck No, that is correct. There may be other equivalent forms.

Answer 13

\[\sin x -\cos x=\sqrt{2}\sin(x-\frac{\pi}{4})\]

Answer 14

I never seen that before @tkhunny
The other option with cos I could think of would be the same as what I wrote just then but with a + sign and sin becomes a cos.

Answer 15

And there will be no solution as well.

Answer 16

Equivalent forms are equivalent. They don't care how you find them or if anyone has met them, before.

Answer 17

so the answers are no solution and sinx−cosx=2√sin(x−π4) ?

Answer 18

No.

Answer 19

That was another method of finding solutions to your question.

Answer 20

oh ok, so what are the final answers then

Answer 21

@tkhunny They care about how you find them. That's called working out.

Answer 22

And if you say it's correct and equivalent, then good for you, well done.

Answer 23

@Azteck Meh. Humans care. Forms and Solutions don't care. :-)

Answer 24

Lol. True!

Answer 25

sorry im still confused, what are the answers in that case?

Answer 26

In all cases, no matter how you work it, the RANGE of \((\sin(x) - \cos(x))\) is \([-\sqrt{2},\sqrt{2}]\).

Now that we know that, we can see that the RANGE of \((\sin(x) - \cos(x))^{2}\) is \([0,2]\).

A little thought on this result should suggest that it will never take on the value 3. Done. No solution.

Answer 27

thanks so much :)