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

Guest Oct 3, 2018
 #1
avatar+92376 
+1

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

 

 

cool cool cool

CPhill  Oct 3, 2018
 #2
avatar+92376 
+1

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

 

 

cool cool cool

CPhill  Oct 3, 2018
edited by CPhill  Oct 3, 2018
 #5
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0

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
avatar+92376 
0

Yep...small mistake...let me correct that  !!!

 

 

cool cool cool

CPhill  Oct 3, 2018
 #7
avatar+92376 
0

Now corrected

 

 

 

cool cool cool

CPhill  Oct 3, 2018
 #3
avatar+92376 
+1

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 ]

 

 

 

cool cool cool

CPhill  Oct 3, 2018
 #4
avatar+92376 
+1

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

 

 

 

cool cool cool

CPhill  Oct 3, 2018

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