How do you solve 2sin^3X-sin^2X-2SinX+1=0? I don't get how you get the 270.

#2sin^3x-sin^2x-2sinx+1=0# for the interval #[0,2pi)#

#2sin^3x-sin^2x-2sinx+1=0# for the interval #[0,2pi)#

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Sean Share
Feb 9, 2018

Answer:

#x=30^@, 90^@, 150^@, 270^@#

Explanation:

.

#2sin^3x-sin^2x-2sinx+1=0#

Let's factor:

#sin^2x(2sinx-1)-(2sinx-1)=0#

#(sin^2x-1)(2sinx-1)=0#

Set each equal to zero and solve for #sinx#

#sin^2x-1=0#

#sin^2x=1#

#sinx=+-1 :.x=pi/2 or 90^@ and x=(3pi)/2 or 270^@#

#2sinx-1=0#

#2sinx=1#

#sinx=1/2 :.x=pi/6, (5pi)/6 or x=30^@, 150^@#

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dk_ch Share
Feb 9, 2018

#2sin^3x-sin^2x-2sinx+1=0#

#=>sin^2x(2sinx-1)-1(2sinx-1)=0#

#=>(2sinx-1)(sin^2x-1)=0#

So
when #2sinx-1#=0
#sinx=1/2=sin(pi/6)=sin(pi-pi/6)#

Hence #x=pi/6,(5pi)/6#
when
#sin^2x-1=0=>sinx=pm1#

when
#sinx =1=sin(pi/2)#
#=>x=pi/2#

Again

when
#sinx =-1=sin(pi+pi/2)#

#=>x=(3pi)/2#

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Feb 9, 2018

Answer:

Factorize

Explanation:

#2sin^3x-sin^2x-2sinx+1 = 0#
#sin^2x(2sinx-1)-1(2sinx-1) = 0#
#(sin^2x-1)(2sinx-1) = 0#
#sin^2x = 1# or #2sinx = 1#
#sinx = +-1# or #sinx = 1/2#
#x = pi/2,(3pi)/2# or #x = pi/6,5pi/6#

#(3pi)/2# is #270^o#

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