Question #db34a

Question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-25T14:17:52

$x=pi/6 and (5pi)/6$ and $x=(3pi)/2$

$cos2x = sinx$

Substitute using the trig identity $cos2x = 1-2sin^2x$

$1-2sin^2x=sinx$

Subtract $1-2sin^2x$ from both sides.

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

Factor

$0=(2sinx-1)(sinx+1)$

Set each factor equal to zero and solve.

$2sinx-1 = 0$ and $sinx+1=0$

$sinx=1/2$ and $sinx=-1$

From the unit circle for $0<=x<2pi$

$x=pi/6 and (5pi)/6$ and $x=(3pi)/2$

Answer 2 · 2017-05-25T14:22:06

The solutions are $S={pi/6+2kpi,5/6pi+2kpi,3/2pi+2kpi}$ , $k inZZ$

We need

$cos2x=1-2sin^2x$

Therefore,

our equation becomes

$cos2x=sinx$

$1-2sin^2x=sinx$

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

We solve this like a quadratic equation

The discriminant is

$Delta=b^2-4ac=1-4*(2)*(-1)=1+8=9$

As,

$Delta>0$ , there are 2 real solutions

$sinx=(-b+sqrtDelta)/(2a)=(-1+sqrt9)/(2*2)=2/(4)=1/2$

$x =pi/6+2kpi$ and $x=5/6pi+2kpi$ , $k inZZ$

and

$sinx=(-b-sqrtDelta)/(2a)=(-1-sqrt9)/(2*2)=-4/(4)=-1$

$x=3/2pi+2kpi$ , $k inZZ$