Question

Find all the solutions in the interval [0,2pi) for cos2x= 1/cos^2x - sin^2x

Answer 1

\[\cos2x = 1\div \cos ^{2}x-\sin ^{2}x\]

here is the equation again

Answer 2

does it help to know that
\[\cos(2x)=\cos^2(x)-\sin^2(x)\]?

Answer 3

i know that its the double angle formula

Answer 4

@bridgetx516x \[\cos 2x=\cos^2 x-\sin^2 x\]

Answer 5

so you have
\[\cos^2(x)-\sin^2(x)=\frac{1}{\cos^2(x)-\sin^2(x)}\]

Answer 6

yes but i didnt know if i should bring it all to one side or multiply by the denominator

Answer 7

the only way for a number to be equal to its reciprocal is if it is \(1\) or \(-1\)

Answer 8

okay

Answer 9

so instead of multiplying a bunch of stuff out, you know
\[\cos^2(x)-\sin^2(x)=1\] or
\[\cos^2(x)-\sin^2(x)=-1\]

Answer 10

oh i wasnt sure if cos2(x)−sin2(x)=−1 but that makes sense now

Answer 11

so either sine is zero and cosine is one, or cosine is zero and sine is one

Answer 12

i think cosine is zero and sine is one

Answer 13

sure for example at \(\frac{\pi}{2}\)
you have
\[\cos^2(\frac{\pi}{2})-\sin^2(\frac{\pi}{2})=-1\]

Answer 14

but also at \(0\) you have
\[\cos^2(0)-\sin^2(0)=1-0=1\] so that one works as well

Answer 15

so would my final answer be 0, pi/2, pi, 3pi/2

Answer 16

\[\cos^2 2x=1\] which means cos 2x=\[\pm1\] in[0,2pi) there are 2 solution one 0 and another pi

Answer 17

also pi/2 and 3*pi/2

Answer 18

4 solutns whole

Answer 19

yes for solutions, you have them
you can check that they work

Answer 20

btw, even if you multiplied both sides by cos^2 - sin^2
\[ \cos^2(x)-\sin^2(x)=\frac{1}{\cos^2(x)-\sin^2(x)} \]
you would get
\[ (\cos^2(x)-\sin^2(x))^2 = 1\]
take the square root, and you get
\[ \cos^2(x)-\sin^2(x) = ±1\]
just as satellite said you would

at this point , if you replaced cos^2 with 1- sin^2 you can solve for sin(x)
if you did not notice that 0, and ±1 for sin and cos are the solutions

Answer 21

than you everyone!