Question

sin^4x-cos^4x=1/2

cos5x=cos(x-pi/3)

cos2x+sin2x+1=0

Answer 1

a)

sin ^4 x + cos ^4 x = 1/2

(sin ^2 x  - cos ^2 x)*( sin ^2 x + cos ^2 x) = 1/2

(sin ^2 x - cos ^2 x) * 1 = 1/2

sin ^2 x - cos ^2 x = 1/2

sin ^2 x - [ 1 - sin ^2 x] = 1/2

2 sin ^2 x = 1 + 1/2

2 sin ^2 x = 3/2    / :2

sin^2 x = 3/4

sin x = - p(3)/2   lub  sin x = p(3)/2

x = pi/3 + 2 pi *k

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lub

x = - pi/3 + 2 pi *k , gdzie  k   liczba całkowita

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b)

cos 5x = cos ( x - pi/3)

5x = x - pi/3

5x - x = - pi/3

4x = -pi/3  + 2 pi *k    / : 4

x = - pi/12 + (1/2)*pi *k,  k - liczba całkowita

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c)

cos 2x + sin 2x + 1 = 0

[ cos ^2 x - sin ^2 x] + [2 sin x * cos x ] + [ sin^2 x + cos ^2 x] = 0

2 cos ^2 x + 2 sin x * cos x = 0

2 cos x * [ cos x + sin x ] = 0

cos x = 0  lub  cos x + sin x = 0

x = pi/2 + pi *k

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sin x = - cos x  / : cos x

tg x = - 1

x = -pi/4 + pi *k , gdzie k  liczba całkowita

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