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# Solution of sin(x)=1-2cos^2(x)

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"Simplify using identities so that you can solve the equation. Give the general solution in radians."
sin(x)=1-2cos^2(x)

I ended up getting...
x = - pi/6 + n2pi, x = - 5pi/6 + n2pi, x = pi/2 + n2pi
...which apparently is incorrect. I'd appreciate some help here since I'm not sure how else to solve this.

Jan 16, 2018

#1
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sin (x)  =   1  - 2cos^2 (x)

Note  that   cos^2(x)   =  1 - sin^2(x)

So  we have

sin x  =   1 - 2  [  1  - sin^2 (x) ]

sin x  =  1  - 2  +  2sin^2(x)

sin (x)  =  - 1  +  2sin^2(x)         rearrange as

2sin^2(x)  -  sin (x)  - 1  =  0        factor as

(2sin (x)  + 1) (sin (x) - 1)  =  0

Se each factor to 0 and solve

2sin x  + 1  =  0      subtract  1 from  both sides

2sin x  = -1         divide both sides by 2

sinx  =  -1/2    and this occurs  at    7 pi / 6    + n*2pi   and  11pi / 6  + n*2pi

[ where n is an integer]

And..for the other solution we have that

sin x  - 1   = 0   add 1 to both sides

sin x  = 1     and this occurs at   pi/2 + n*2pi   Jan 16, 2018
#2
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Sweet, thanks for showing all the steps. Really helps me understand.

Aleguan  Jan 16, 2018
#3
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You're welcome  !!!!   Jan 16, 2018