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# Trigonometry Questions

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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
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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   Apr 29, 2019
#2
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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   Apr 29, 2019
#3
+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   Apr 30, 2019