Question

Find the exact solutions of sin^2X + Sin2X = -Cos^2X + 1 in the interval [0, 2pi)

Answer 1

Hey did you try putting everything on one side

Answer 2

And then us
\[\cos^2(x)=1-\sin^2(x)\]

Answer 3

use*

Answer 4

No.... I didn't try that. But I have no idea how to do it. can you take it from the top...

Answer 5

Put everything on one side

Answer 6

add cos^2(x) on both sides
subtract 1 on both sides

Answer 7

yeh, I have it all on 1 side.

Answer 8

ok the use that identity I mentioned and replace cos^2(x) with 1-sin^2(x)
tell me if anything cancels :)

Answer 9

Got it

Answer 10

I got the pi/2 and 3pi/2 but there's two more answers (0 and pi) which I don't get how they got.

Answer 11

you just have to write the equation like that
sin^2(x)+cos^2(x)+sin(2x)=1

Answer 12

I got to sin2X=0 then solved for x= pi/2 +2 n pi and X=pi + 2 n pi

Answer 13

\[0 \le x < 2 \pi\]
\[\sin(2x)=0\]
if we let u=2x then we have
\[\sin(u)=0 \text{ where } 0 \le \frac{u}{2} < 2 \pi \]
\[\sin(u)=0 \text{ where} 0 \le u < 4 \pi\]
What are values of u that satisfy this equation between 0 and 4 pi include 0

Answer 14

Oh... Okay. How are you making the sign for pi and x? (For future reference)