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barxatty [35]
3 years ago
7

Determine the general solution of: sin 3x - sin x = cos2x

Mathematics
1 answer:
SpyIntel [72]3 years ago
6 0
Sin³ x-sin x=cos ² x

we know that:
sin²x + cos²x=1  ⇒cos²x=1-sin²x
Therefore:

sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0

sin³x=z

z³+z²-z-1=0

we divide by Ruffini method:
              1     1     -1     -1
        1           1      2      1                z=1
-------------------------------------
              1     2      1      0
       -1         -1      -1                       z=-1
--------------------------------------
              1     1       0                       z=-1

Therefore; the solutions are z=-1 and z=1

The solutions are:
if z=-1, then
sin x=-1   ⇒x= arcsin -1=π+2kπ    (180º+360ºK)   K∈Z


if z=1, then
sin x=1   ⇒ x=arcsin 1=π/2 + 2kπ   (90º+360ºK)   k∈Z

π/2 + 2kπ    U   π+2Kπ=π/2+kπ     k∈Z    ≈(90º+180ºK)

Answer: π/2 + Kπ    or     90º+180ºK          K∈Z
Z=...-3,-2,-1,0,1,2,3,4....
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