Question

giving a medal!
find all solutions in the interval(0,2pi)

sin^x-cos^x=0

Answer 1

Need clarification. What do you mean by " sin^x "?

Answer 2

oops,sorry:)
it is:
sin^2x-cos^2x=0

Answer 3

or even better, (sin x)^2 - (cos x)^2 = 0. What have you thought of doing so far?

Answer 4

i tried using the power reducing formulas and substituting them in this equation.

Answer 5

am i on the right track?

Answer 6

You might find it easier to use a substitution for either (sin x)^2 or (cos x)^2.

Which trig identity would apply nicely here?

Answer 7

pythagorean trig identity?
sin^2x+cos^2x=1

Answer 8

@mathmale ?

Answer 9

that's be neat. from that identity you could obtain (cos x)^2 = 1 - (sin x)^2.

Want to try that? Doing so will eliminate the (cos x)^2 from your equation and leave you with (cos x)^2 alone.

Answer 10

ok:
sin^2x-(1-sin^2x)=0
sin^2x-1+sin^2x=0
2sin^2x-1=0

Answer 11

Yes, and then 2 (sin x)^2 = 1, and (sin x)^2 = 1/2. What next? There will be more than one solution on the given interval.

Answer 12

so,the nest step is:
sin x=square root of 1/2
is this right?

Answer 13

Yes, except that you need to write sin x = plus or minus 1/Sqrt(2).

Answer 14

so,what r the solutions?

Answer 15

x=pi/4,3pi/4,5pi/4,7pi/4?

Answer 16

does anyone want to help me?

Answer 17

@mathmate ?

Answer 18

anybody there??????

Answer 19

Your answer is correct!

Answer 20

Oh,thank god.Thankyou very much:)

Answer 21

You are welcome :)