Question

How do you do this?

Answer 1

i am about to attach a file sorry

Answer 2

@gorv

Answer 3

Can you do first part?

Answer 4

going from sinxcosx=1 to sin^4x-sin^2x+1=0? If thats what you are talking about then yes

Answer 5

Ok good, so now you have \(\sin^4 x-\sin^2 x+1 = 0\)

Hmm, it looks like quadratic equation... You can let \(u\) be \(\sin^2 x\)

So now you have \(u^2-u+1=0\)

Apply quadratic formula.

Answer 6

ok give me one sec to do this

Answer 7

ok so i got\[(\sin ^2x \pm \sqrt{\sin ^2x(-1-4\sin^2x)})/2\sin^4x\]

Answer 8

i don't know where to go from here

Answer 9

@geerky42

Answer 10

please help nnesha

Answer 11

?

You just solve for u, you apply formula to solve for u; you have

\(u = \dfrac{1\pm\sqrt{1-4}}{2} = \dfrac{1\pm i\sqrt{3}}{2}\)


Notice that \(u = \sin^2 x = \dfrac{1\pm\mathbf i\sqrt{3}}{2}\) is imaginary number.