Question

Solve for x in [0,2pi]
Cos^2x-sin^2x=0

Answer 1

Put it in tangent form then solve it using the range given

Answer 2

How do I put it in tangent form?

Answer 3

Like this,
\[\cos^2(x)-\sin^2(x)=0\]
\[\cos^2(x)=\sin^2(x)\]
\[1=\frac{\sin^2(x)}{\cos^2(x)} \]
\[ \tan^2(x)=1\]
\[\tan(x)=\pm 1\]

Answer 4

Now we have "\(\pm \)" so we'll use all the quadrants

Answer 5

?? \(\cos^{2}(x) - \sin^{2}(x) = \cos(2x)\).

Why do it some hard way?

Answer 6

\[\tan(x)=1 \\ x=45 \]
\[x=180-45=135 \\ x=180-45=225 \\ x=360-45=315\]
4 answers

Answer 7

Awesome, thank you so much. @.Sam.

Answer 8

Unfortunately, the "tangent" method introduced Domain restrictions. If cos(x) = 0 had been a solution to the original equation, you would have discarded it with the division to tangent. Always keep the Domain in mind. Always worry about changing it.

Answer 9

How would I do sin3x+sinx=0 with the same domain?

Answer 10

sin(3x)+sin(x)=0?

Answer 11

Yes

Answer 12

Looks horrible... thankfully, we have our ever useful identities... unless you already know what sin(3x) is...

Do you, by any chance? :D

Answer 13

No

Answer 14

@terenzreignz

Answer 15

Well, two things... first
\[\Large \sin(\alpha + \beta)= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\]
and
\[\Large \sin(3x) = \sin(2x+x)\]

Answer 16

\(\sin(3x) + \sin(x) = \sin(2x + x) + \sin(2x - x)\) might lead to something.

You just have to play with it. You won't break it. Just try something.

Answer 17

Ok @ terenzreignz

Answer 18

Okay?
So expand
sin(3x)
using the fact that
sin(3x) = sin(2x + x)

and
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

^_^

Answer 19

What are a and b?

Answer 20

@terenzreignz

Answer 21

Nothing, this is just a formula.
If you want, you can have a = 2x and b = x
and work from there...

Answer 22

Ok, thank you so much.

Answer 23

No problem :)