Question #db34a

Question

1295 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#