How do you solve $cos 2 x - sin x = 1$ $cos2x-sinx=1$ in the interval [0,360]?

Question

21977 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T20:23:46

The answer is $S={0,180,210,330}º$

We use

$cos2x=1-2sin^2x$

The equation becomes

$1-2sin^2x-sinx=1$

$2sin^2x+sinx=0$

$sinx(2sinx+1)=0$

So, $sinx=0$ or $sinx=-1/2$

$x=0, 180$ º and $x=210, 330$ º

How do you solve cos2x-sinx=1cos2x−sinx=1cos2x-sinx=1 in the interval [0,360]?