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Find all real numbers in the interval [ 0 , 2pi ) that satisfy each equation. Round approximate answers to the nearest tenth.

 

1. tan(-x) = tan(x)

 

2. 2 cos^2(x) = 3 cos(x)

 

3. tan (x) = sec x - sqrt(3)

 

4. 5 sin^2(x) - 2 sin(x) = cos^2(x)

 

 

Thanks.

 Apr 29, 2019
 #1
avatar+111433 
+2

1. tan(-x) = tan(x)

 

Note that    tan (-x)  = -tan(x)....so we have

 

-tan (x)  = tan(x)          add tan (x) to both sides

 

0 =  2 tan x               divide both sides by 2

 

0  = tan x   ....   and this occurs at x = 0   and x  = pi

 

 

cool cool  cool

 Apr 29, 2019
 #2
avatar+111433 
+2

2.   2 cos^2(x) = 3 cos(x)     subtract  3cos (x) from both sides

 

2cos^2(x)  - 3 cos (x)  = 0       

 

cos (x)  [  2cosx - 3 ]  = 0

 

So

 

cos x  = 0.....and this occurs at  x=    pi/2   and   x = 3pi/2

 

Or

 

2cosx - 3  = 0

2cosx  = 3

cos x = 3/2     impossible

 

 

 

cool cool cool

 Apr 29, 2019
 #3
avatar+111433 
+2

4. 5 sin^2(x) - 2 sin(x) = cos^2(x)

 

5sin^2 x - 2sin x  = 1 - sin^2 x

 

6sin^2x - 2sin x - 1  =  0

 

(3sin x + 1) ( 2sin x - 1)  = 0

 

3sin x  = 1

sin x  = 1/3     take the arcsin

arcsin (1/3) =   x   ≈  .34 rads      and x ≈  [ pi - .34] rads ≈  2.80 rads

 

And

 

2sin x  - 1  = 0

2sin x  = 1

sin x  = 1/2      and this occurs at x = pi/6    and x  = 5pi/6

 

 

cool cool cool

 Apr 30, 2019

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