Solving For x Over An Interval

[tristatefabricatorsinc]

Solve for x over the interval [0 degrees, 360 degrees];
2cosx sinx - cosx = 0

Does anyone know where to begin on this? I think I ma supposed to treat it liek a quadratic, so if I try to break it down I can get as far as

cosx sinx = 0

Thanks

 

[Guest]

2cosx sinx - cosx = 0

cosx (2 sinx - 1) = 0

then either cos x = 0
or 2 sinx -1 = 0
...
2 sinx = 1
sinx = 1/2

solve these 2 for your answers.
cos x = 0
sinx = 1/2

 

[tristatefabricatorsinc]

For cos x = 0
I got 90 degrees, and 270 degrees

For Sin x
I got 30 degrees and 210 degrees

Am I correct?

 

[tristatefabricatorsinc]

Can someone tell me if I was correct with my answers above?

Thank You Very Much For All The Help!

 

[soroban]

Hello, tristatefabricatorsinc!

For $\displaystyle \cos x\,=\,0$, I got: 90^o, 270^o.

For $\displaystyle \sin x \,-\,1\:=\:0$, I got: 30^o, 210^o . . . no

Sine is negative in quadrant 3 . . . Your second angle should be 150^o.