Question

i couldn't find the answer for this question, not sure if i'm remotely close
the problem says solve cos 2x=2(sinx-1) for all real x.

Answer 1

cos 2x can be expressed in terms of sin^2(x), right? If you do that, you get a quadratic in sin (x) which can be solved. Try it!

Answer 2

so next step is sin^2 (x) = 2sinx - 2 ?

Answer 3

How do we express cos(a+b)?

Answer 4

cos(a+b)=cos(a).cos(b)-sin(a).sin(b), remember?

Answer 5

wait i remember that, but cos2x in the beginning can turn into 1-2sin^2? right

Answer 6

Exactly!

Answer 7

So if you think of sin^2(x) as some variable, say y, we get a quadratic equation in y which can be solved.

Answer 8

1-2sin^2(x)= 2(sinx - 1)

Answer 9

Yes right.

Answer 10

so then i try to express it in a way that it looks like a^2 + a value - a value and then factor?

Answer 11

Yes!

Answer 12

not sure if this is right, 2sin^2 (x) -2sinx-1=0

Answer 13

Let me check

Answer 14

Lets work it out:
\[\cos(2x)=1-2\sin^{2}(x)\]
So:
\[1-2\sin ^{2}(x)=2(\sin(x)-1)=2\sin(x)-2\]

Answer 15

Gives me:
\[2\sin ^{2}(x)+2\sin(x)-3=0\]

Answer 16

Right??

Answer 17

oh, i see where i went wrong

Answer 18

Now on it is the application of the quadratic formula to get the value of sin(x) and hence (x)

Answer 19

okay, and then set them to zero meaning (first equation)=0 (second equation) = 0

Answer 20

Not sure what you mean. I would substitute sin(x) by another variable, say y to get a quadratic as:\[2y ^{2}+2y-3=0\]

Answer 21

The solve for y

Answer 22

Got it?