Question

Find all solution in the interval (o, 2pi]

Answer 1

\[\sin^2 x- \cos^2 x=0\]

Answer 2

there are several ways,
one of them is to use \(\sin^2x+\cos^2x=1\)
so, plug in \(1-\sin^2x\) in place of cos^2 x in your equation.
what u get ?

Answer 3

\[2 \sin^2 x=2\]?

Answer 4

do you recognize \(\sin^2(x)-\cos^2(x)\) as something else?

Answer 5

no, where does 2 come from ?
did you mean
2 sin^2 x = 1 ?

Answer 6

Yes, 2 sin^2 x = 1

Answer 7

if you know the double angle formula for cos (cos 2x =....?) then this would get easier.

Answer 8

ok, lets continue with that,
isolate sin x then,
sin x =... ?

Answer 9

\[\sin x= \sqrt{1/2}\] ?

Answer 10

thats a positive root,
what about negative root.
\(a^2=b \implies a=\pm \sqrt b\)

Answer 11

\[-\sqrt{1/2}\] then?

Answer 12

they both
\(\sin x = 1/\sqrt2 \quad \sin x=-1/\sqrt2\)
do you know sin of what angle equals those ?

Answer 13

\[\pi/4 & 5\pi/4 \]

Answer 14

pi/4 & 5pi/4

Answer 15

correct!

Answer 16

Thanks you!!! :D

Answer 17

welcome ^_^