How would I simplify 1-2sinx=-cosx to the point where I could solve for x on [0,2pi)

Question

amistre64 · Answer

1 = sin^2+cos^2

amistre64 · Answer

1-2sinx=-cosx

 sin^2+cos^2 -2sin = -cos

(sin^2-2sin) = - (cos^2 +cos)

anonymous · Answer

I got sinx(sinx-2)=-cosx(cosx+1)

anonymous · Answer

I don't see where I could go from here except for maybe getting a -tanx.

amistre64 · Answer

one idea is to complete the squares

sin^2-2sin = - (cos^2 +cos)

sin^2-2sin +1 -1 = - (cos^2 +cos +1/4) + 1/4

(sin -1)^2 -1 = - (cos + 1/2)^2 + 1/4

(sin -1)^2 + (cos + 1/2)^2 = 5/4

$\frac{4}{5}$[(sin -1)^2 + (cos + 1/2)^2] = 1

jsut a thought tho :)

amistre64 · Answer

think of a circle centered at (1,-1/2) with a radius of sqrt(5)/2

anonymous · Answer

?

anonymous · Answer

Can you put hat into the equation editor thingy please?

amistre64 · Answer

$$(\sin x -1)^2 + (\cos x + 1/2)^2 = \frac54$$

anonymous · Answer

Wait, but how does that get me closer to solving for x?

amistre64 · Answer

let u = sin(x), let v = cos(x)  but maybe you want a different method

anonymous · Answer

:? I'm so lost now.

anonymous · Answer

@cwrw238