Question

sin^2x-cos^2x=0

Answer 1

Soyou know what \(\sin^2(x)\) and \(\cos^2(x)\ can be rewritten as?

Answer 2

Do you*

Answer 3

Recall the identity that sin^2x+cos^2x = 1 so sin^2 x = 1 - cos^2 x.

Answer 4

sin^2x-cos^2x=0

1 - cos^2 x - cos^2 x = 0

Answer 5

So with what @Directrix mentioned, you can see that the sine and cosine functions are both positive. How do we change \(-\cos^2(x)\) to be positive?

Answer 6

you could also simply divide cos^2x both sides to get

tan^2x = 1

Answer 7

@youhan_teh please note that:
\[\cos(2x)=(\cos x)^{2}-(\sin x)^{2}\]

Answer 8

@youhan_teh so your equation, after substitution, becomes:
\[\cos(2x)=0\]

Answer 9

I havent seen @youhan_teh respond once.

Answer 10

1- 2 cos^2 x = 0

-2 cos^x = -1

cos^2 x = 1/2

cos x = +/- sqr(1/2)

cosx = +/- sqr(2) / 2

Answer 11

so then we use duble angle identity ?
sorr, new here. im trying to figure this out

Answer 12

Oh and welcome to openstudy!! :)

Answer 13

Thank you,

Answer 14

slvin the rewritten equation, you will get:
\[2x=\frac{ \pi }{ 2 }+2k \pi\]
where k is an integer with sign, zero is included.
now please divide by 2 the above solution, and you will get your answer

Answer 15

@Michele_Laino im not sure how you got there.

Answer 16

@youhan_teh sorry:
not
\[2k \pi\]
but
\[k \pi\]

Answer 17

because, cos(y) is zero when y=90°, 270°, 450°...an so on, namely:
\[y=90°+k*180°\]

Answer 18

Thank you. i think im getting this. :)

Answer 19

now, in your case we have: y=2x, so we can write:
\[2x=\frac{ \pi }{ 2 }+k*\pi\]
finally dividing by 2, we\[x=\frac{ \pi }{ 4 }+k*\frac{ \pi }{ 2 }\] have:

Answer 20

Thank you!