Question

como resolver em IR esta equação
$sin(x) - cos(x) = ( \sqrt{6} ) \div 2$
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Answer 1

Resposta:

S = { x ∈ IR/ x = 7π/12 + 2kπ ou x = 11π/12 + 2kπ, k ∈ Z}

Explicação passo-a-passo:

senx - cosx = √6/2

senx - sen(π/2 - x) = √6/2

2sen([x -(π/2 - x)]/2cos(x + π/2 - x)/2 = √6/2

2sen(x - π/2 + x)/2cosπ/4 = √6/2

2sen(x - π/4).√2/2 = √2.√3/2

2sen(x - π/4) = √2.√3/2 . 2/√2

2sen(x - π/4) = √3

sen(x - π/4) = √3/2

sen(x - π/4) = senπ/3

x - π/4 = π/3 + 2kπ

x = π/4 + π/3 + 2kπ

x = 7π/12 + 2kπ

ou

x - π/4 = π - π/3 + 2kπ

x = π/4 + π - π/3 + 2kπ

x = (3π + 12π - 4π)/12 + 2kπ

x = 11π/12 + 2kπ

S = { x ∈ IR/ x = 7π/12 + 2kπ ou x = 11π/12 + 2kπ, k ∈ Z}