Question

Question #db34a

Answer 1

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

Explanation:

`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

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

Explanation:

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`