Solve the following equation cos22x sin2 2x 0 sin2x sin x

Solve the following equation cos^2(2x) - sin^2 (2x) = 0 sin(2x) = sin x

Solution

13.

cos²(2x)-sin²(2x)=0

(cos(2*2x))=0

cos(4x) =0

in interval (0,2):

4x =_/2 ,3_/2

x =_/8 ,3_/8

in general:

cos(4x) =0

4x =(2n+1)_/2

x =(2n+1)_/8 where n is an integer

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14)sin(2x)=sinx

2sinxcosx =sinx

2sinxcosx -sinx =0

sinx(2cosx-1)=0

sinx=0, 2cosx-1=0

sinx=0,cosx=1/2

in interval [0,2):

sinx=0

=> x=0,

cosx=1/2

=>x=/3,5/3

in general:

sinx=0

=>x=n

cosx =1_/2

=>x=(2n -(_/3)),(2n+(_/3)) where n is an integer