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Solve for x on the interval [0,2pi]

Guest Oct 30, 2018
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15sin^2 (2x)  = 1 + sin (2x)

                        _________          multiply both sides by 2

                              2

 

30sin^2(2x)  =  1 + sin(2x)     rearrange  as

 

30sin^2(2x) - sin(2x) - 1   =  0       factor

 

(6 sin (2x)  + 1)  ( 5sin (2x) - 1)  = 0

 

Set each factor to 0  and solve for x

 

6sin(2x) + 1   =  0

6sin(2x)  = -1

sin(2x)  = -1/6

sin(x)  = -1/6  at  ≈ 3.309 rads , 6.12 rads, 9.59 rads  and 12.40 rads

So dividing each of these by 2 we get

x ≈ 1.6545 rads, 3.06 rads, 4.795 rads and 6.20 rads

 

And

5sin (2x)  - 1  = 0

5sin (2x) = 1

sin(2x) = 1/5   at  ≈ .201 rads, 2.94 rads, 6.48 rads, 9.22 rads

Divide both of these by 2 and we get

x ≈ .105 rads, 1.97 rads, 3.24 rads and 4.61rads

 

 

 

cool cool cool

CPhill  Oct 31, 2018

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