Question

solve for θ. sin^2 θ= cos^2 θ

Answer 1

so \(\sin(\theta)=\cos(\theta)\) or
\(\sin(\theta)=-\cos(\theta)\)
try\(\frac{\pi}{4},\frac{3\pi}{4}...\)

Answer 2

I'm a bit confused isn't it
1/2(1-cos ^2θ) = 1/2(1+ cos^ θ)

Answer 3

\[a^2=b^2\implies a=b\text{ or }a=-b\]

Answer 4

\[\sin^2\theta+\cos ^{2}\theta = 1\]

Answer 5

maybe i am reading it wrong
thought it was
\[\sin^2(\theta)=\cos^2(\theta)\]

Answer 6

yup

Answer 7

Yes

Answer 8

ok then my answer above should be correct

Answer 9

\[\frac{\pi}{4}\] they are the same, also at \(\frac{5\pi}{4}\)
at \(\frac{3\pi}{4}\) one is the negative of the other, also at \(\frac{7\pi}{4}\)

Answer 10

Divide both sides by cos^2(x)

tan^2(x)=1
tan(x) = +1, -1

Answer 11

I got + and - pi/4

Answer 12

all of these answers are probably correct, just a lot of different ways to solve this problem

Answer 13

there are an infinite number of answers, but in the interval \([0,2\pi)\) there are 4

Answer 14

yeah, i see that now, good job