Can the equation 2 - cos2x =sinx be solved within a specific interval?

[FlopperJr]

Homework Statement

Solve the equation 2 - cos²x =sinx. Give the solution in the interval 0≤x≤360.

Homework Equations

sin²x+cos²x=1
I know for sin you take inverse. Then subtract that from 180. I believe those are basic angle then you +/- 360. ?

The Attempt at a Solution

I tried to manipulate it to work and solve but I am not sure what else and then how to.
2 - cos²x =sinx
2(1-cos²x)=sinx
2(sin²x)=sinx

 

[rollcast]

Its too late at night for me to start working that out but have you tried the trig identity

tanx=sinx/cosx

 

[FlopperJr]

Um, i think so. Wait yes i did.

 

[Clever-Name]

I would suggest writing it in a form that's quadratic in sin(x). Use the squared identity but in the other way.

edit - also, this step:2(1-cos^{2}(x))=sin(x) is NOT valid. Can you see why?

 

[SammyS]

Homework Statement

Solve the equation 2 - cos²x =sinx. Give the solution in the interval 0≤x≤360.

Homework Equations

sin²x+cos²x=1
I know for sin you take inverse. Then subtract that from 180. I believe those are basic angle then you +/- 360. ?

The Attempt at a Solution

I tried to manipulate it to work and solve but I am not sure what else and then how to.
2 - cos²x =sinx
2(1-cos²x)=sinx
2(sin²x)=sinx

The following two equations are not equivalent.

 2 - cos²x = sinx    &     2(1-cos²x) = sinx

The first equation is equivalent to

1 + (1 - cos²x) = sinx​

Now substitute sin²x for 1 - cos²x and then subtract sinx from both sides. You then have a quadratic equation in sinx .

 

[Curious3141]

Of course, the quadratic may not have real roots (and doesn't in this case). This is a pickle if you want real solutions for x.

(To be less cryptic, I meant that there are no real solutions here).