How do I solve on the interval from [0, 2pi): 2cos^x-sinx-1=0
??????? HELP PLEASE

Question

How do I solve on the interval from [0, 2pi): 2cos^x-sinx-1=0 
??????? HELP PLEASE

anonymous · Answer

Write it out in latex please :)

anonymous · Answer

$$2\cos ^{2}x-\sin x-1=0$$

saifoo.khan · Answer

Use the identity:
$$\sin^2x + \cos^2x =1$$

anonymous · Answer

I have been trying that, but I keep getting stuck

saifoo.khan · Answer

2(1-sin^2x) - sinx -1 =0

2 - 2sin^2x - sinx - 1 =0

-2sin^2x - sinx +1 =0

Now suppose sinx as y.

Rewrite:

-2y^2 - y + 1 = 0

Solve the quadratic.

anonymous · Answer

But I'm solving on an interval of 0 to 2pi, so it needs to be answers like pi/4 pi/6 etc...

saifoo.khan · Answer

Yes. Once you get the value of y, place them in the equation you made:
sinx = y.

Solve for y.

anonymous · Answer

My teacher never taught us to use y...so now I'm confused out of my mind...

anonymous · Answer

we're working with the unit circle

saifoo.khan · Answer

Alright then. Factor this out:
-2sin^2x - sinx +1 =0

anonymous · Answer

would you factor out negative one?

saifoo.khan · Answer

You can do that too. Your answer will be same.

mertsj · Answer

$$-2\sin ^2x-\sin x+1=0$$

mertsj · Answer

$$2\sin ^2x+\sin x-1=0$$

mertsj · Answer

$$(2\sin x-1)(\sin x+1)=0$$

mertsj · Answer

Set each factor equal to 0 and solve.