Question

2sinx -2cosx = √2, solve for x.

ANSWER: 5π/12 and 13π/12

Answer 1

divide by sqrt 2 on both sides
what u get?

Answer 2

Haha, that'll give me 2/√2(sinx) - 2/√2(cosx) = 1

Answer 3

Then I'd factor √2/2 out and eventually end up with the same wrong answer I've been getting...

Answer 4

sorry, 2/√2, not √2/2

Answer 5

note that 1/sqrt 2 = sin 45 = cos 45
so u have
cos 45 sin x - sin 45 cos x
you remember sin(A_B) formula ?

Answer 6

Haha, it isn't 1/√2 though, it's 2/√2?

Answer 7

Ah, but I guess that's the same as 2(1/√2)...so then is it 2(sin45-cos45)=1?

Answer 8

yeah, youu can factor that 2 out

Answer 9

where did 'x' go ? :O

Answer 10

Haha, yeah that's what I was wondering.

Answer 11

xD

Answer 12

sin A cos B - cos A sin B = sin (A-B)
cos 45 sin x - sin 45 cos x = sin (x-45)
makes sense ?

Answer 13

Not at all.

Answer 14

did u know this ?
sin A cos B - cos A sin B = sin (A-B)

Answer 15

Why are you introducing sin and cos of 45 degrees?
Yeah, I know the double angle identities.

Answer 16

Or compound, whichever one this is.

Answer 17

to simplify the left side
1/sqrt 2 = sin 45 = cos 45
that simplifies left side greatly

Answer 18

This doesn't make sense. Thanks for trying, but it's only confusing me further.

Answer 19

so did u not get this ?

sin A cos B - cos A sin B = sin (A-B)
sin x cos 45 - cos x sin 45 = sin (x-45)

Answer 20

@KinzaN , @hartnn has a good approach

Answer 21

Is there anything that he said you can point out that doesn't make sense so I can clarify it or so he can?

Answer 22

\[2 ( \sin(x)-\cos(x))=\sqrt{2} \]
\[\frac{1}{\sqrt{2}} \sin(x)-\frac{1}{\sqrt{2}} \cos(x)=\frac{1}{2}\]
Have you gotten this far?

Answer 23

Yeah, but beyond that I get stuck after getting an answer of 5π/12--
and I also don't understand why this is the only method that works. There must be another that's simpler which we're expected to use.

Answer 24

Sine is positive in the first and second quadrants, so why is their answer of 13π/12 in the third? Wah.

Answer 25

Ah, never mind! I got both solutions using this method. TYTY.