Question

Solve sin2x = sqaureroot 2 cos x

Answer 1

left side sin2x = 2sinx coax . square both side, you can have the solution

Answer 2

Is it: \(\sin2x=\sqrt{2}\cos x\), or : \(\sin2x=\sqrt{2\cos x}\)?

Answer 3

uhh sry its sin2x=\[\sqrt{2}\] cosx

Answer 4

OK, so if you follow @Hoa's suggestion, you get:
\[(2\sin x \cos x)^2=2 \cos^2x\]Can you do the next step?

Answer 5

:) yes, i can hihi

Answer 6

@Hoa: you an I can, but can @alexprip?

Answer 7

Well I'm not exactly sure I have to find a way to remove the sin so I can have a quadratic equation right?

Answer 8

YOu can split up into two different equations: one with cos²x and one with sin²x.

Answer 9

\(4\sin^2x\cos^2x=2\cos^2x\) can be written as:\[4\sin^2x=2 \vee \cos^2=0\]

Answer 10

That is because it is of this standard type: ab=ac <=> a=0 v b=c

Answer 11

I'm confused..

Answer 12

This one does it this way: http://answers.yahoo.com/question/index?qid=20090510225256AAJh74l

Answer 13

standard type: ab=ac: there is a common factor on both sides.
If a = 0 then you have 0b=0c, so 0 on both sides: This is ok, so a=0 is a solution.
If a is NOT 0, then you can divide both sides by it, leaving b=c.

That is why, as soon as you have an equation like ab=ac, you can replace it with a=0 v b=c.

Answer 14

Now to use this on your equation:
You have\[4\sin^2x \cos^2x=2\cos^2x \]This is \[\cos^2x \cdot 4\sin^2x = \cos^2 x \cdot 2\]Now you can see the ab=ac here:

a = cos²x, b=4sin²x and c=2.

Answer 15

I see but what do I then have to do next?

Answer 16

OK, so now switch to a=0 v b=c:\[\cos^2x=0 \vee 4\sin^2x=2\]YOu can solve these two equations separately, they are easier than the whole thing!

Answer 17

Do you know how to solve \(\cos^2x=0\)?

Answer 18

Well I have to take a look at the unit circle right?
\[\cos^{2}\] x = 0 => \[\frac{ \pi }{ 2 } and \frac{ 3\pi }{ 2 }\]
right?

Answer 19

Or am I completly wrong?

Answer 20

You are right!

Answer 21

Because cos²x=0 is the same as cosx=0.

Answer 22

Great:) And for sin i divide by 4 and get sin squared = 1/2 ?

Answer 23

Yes, so what should be the next step?

Answer 24

sin squared = 1/2
\[\frac{ \pi }{ 6 } and \frac{ 5\pi }{ 6 }\]
?

Answer 25

And that is about it?

Answer 26

No, you have sin²x=1/2, so \[\sin x = \pm \sqrt{\frac{1}{2}}\pm \frac{1}{2}\sqrt{2}\]So you need to find x's that fit this.

Answer 27

But it's your lucky day! These values \(\pm \frac{1}{2}\sqrt{2}\) belong to x's that are also easy to find in the unit circle...

Answer 28

Here it is: