Question

sin^2x-cos^2x=0
solve for the equation for the interval [0, 2pi)

Answer 1

Therefore:
\[\sin^2(x) = \cos^2(x)\]

Answer 2

\(\large sin^2x-cos^2x=0 \)
\(\large (sinx+cosx)(sinx-cosx)=0 \)

can u do it from here?

Answer 3

my choices all have pi/4 in it
not sure how to get there
please help

Answer 4

just set each factor equal to zero and solve....
for example: sinx + cosx = 0
can you solve this?

Answer 5

if we solve dpalnc's equations, we have:
\[\sin(x) = \cos(x)\]
or:
\[\sin(x) = -\cos(x)\]

Answer 6

for sinx = cosx, what angle give the same sine and cosine???
hint: there are two of 'em.... in [0, 2pi)

Answer 7

\[\frac{ \pi }{ 4 }, \frac{ 3\pi }{4 }, \frac{ 5\pi }{ 4 },\frac{ 7\pi }{ 4 }\]so the answer would be

Answer 8

@dpaInc though this does not relate to the question, how did you make your avatar?

Answer 9

yes