Find all solutions in the interval [0, 2π).

sin^2x - cos^2x = 0

Question

Find all solutions in the interval [0, 2π). 

sin^2x - cos^2x = 0

anonymous · Answer

hhaha funny here are the (6 points)

danjs · Answer

i gave you the biggest clue on your last post

$$\sin^2(a) - (1 - \sin^2(a)) = 0$$

sin^2(a) = 1/2

so
$$\sin(a) = \pm \frac{ 1 }{ \sqrt{2} }$$

danjs · Answer

that is the same as

$$\sin(a) = \pm \frac{ \sqrt{2} }{ 2 }$$

each point on the unit circle is
$$(x,y) = (\cos(a), \sin(a))$$

so the y coordinate is the sin(a) value

look where that is root2/2 or -root(2)/2

danjs · Answer

4 points, 4 angles

naveenbhatia1312 · Answer

this is confusing

danjs · Answer

need to practice the easier stuff first

naveenbhatia1312 · Answer

whats the answer?

danjs · Answer

I used the identity  sin^2 + cos^2 = 1

the other is just algebra to solve for sin(a)

sin(a) is that value for certain angles on the unit circle...

parthkohli · Answer

Just another fun solution. It's akin to saying that $-\cos(2x) = 0$ so $\cos(2x) = 0$.

danjs · Answer

angles on here where y coordinate is that value, there are 4

danjs · Answer

goodluck

naveenbhatia1312 · Answer

@ParthKohli what is the answer?

parthkohli · Answer

How would you solve $\cos(2x) = 0$? In general when is cosine zero?

naveenbhatia1312 · Answer

pi/2 or 3pi/2

naveenbhatia1312 · Answer

@ParthKohli