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)

3 Answers
Manikandan S.
Feb 9, 2018

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

P dilip_k
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

Sean
Feb 9, 2018

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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