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# ​I am utterly confused with solving these for theta or x. I think we use trig identities to find x?

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7 I am utterly confused with solving these for theta or x. I think we use trig identities to find x?

Oct 3, 2018

#1
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a. sin^2x  = 2sin^2(x / 2)

Note  sin^2 (x / 2 )   =  1 - cos x

So we have

sin^2x  = 1 - cos x

1 - cos^2x  = 1 - cos x      subtract 1 from both sides

-cos^2x  = -cos x       multiply through  by  -1

cos^2x  = cosx

cos^2x - cos x  = 0

cosx ( cos x - 1)  = 0

Set each factor to 0  and we have that

cos x  =  0     and this happens at   pi/2  + n*pi    where n is an integer

and

cos x  -  1   = 0

cos x  = 1     and this happens at   0  + n* 2pi    where  n is an integer   Oct 3, 2018
#2
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b. sinx  - cos^2 x - 1   = 0

sin x  -  [ 1 - sin^2x ]  - 1   = 0

sin^2x  + sinx  - 2  = 0     factor

(sin x - 1) ( sin x  + 2)  = 0

Set each factor to 0 and solve for  x

sin x  - 1   = 0

sin x  = 1  and this happens at   pi/2 + n* 2pi   where n is an integer

And

six + 2  =  0   has  no solution   Oct 3, 2018
edited by CPhill  Oct 3, 2018
#5
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Hi! Thank you so much for replying! When you distribute the negative, doesn’t the -1 + -1 =-2? So if

t becomes sin^2x+sinx-2=0?

Guest Oct 3, 2018
#6
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Yep...small mistake...let me correct that  !!!   CPhill  Oct 3, 2018
#7
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Now corrected   CPhill  Oct 3, 2018
#3
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c. 2sin^2 θ  + 3sin θ  = 2

2sin^2 θ + 3sinθ - 2  = 0   factor

(2sinθ - 1) (sin θ + 2)   =  0

Setting each factor to 0....solve for  θ

2sinθ - 1  =  0

2sinθ  = 1

sinθ  = 1/2     and this happens at    pi/6 + n*2pi   and   5pi/6 + n*2pi    where n is an integer

And

sin θ  + 2   =0

sinθ  = -2     [ this has no solution ]   Oct 3, 2018
#4
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d. cos (2x) - cos x  = 0

2cos^2x - 1 - cos x   =0

2cos^2x  - cos x  - 1   =  0   factor

(2cos x + 1) (cos x - 1)  = 0      set each factor to  0  and solve for  x

2cos x  + 1  = 0

2cos x  = -1

cos x  = -1/2    and this happens at   2pi/3 + n*2pi    and  4pi/3 + n*2pi   where  n is an integer

And

cos x - 1   =  0

cos x   = 1    and this happens at   0 + n*2pi    where n is an integer   Oct 3, 2018