Resposta:

3 sen(x) + 4 cos(x) = cos(2x) * [4 cos(x) − 3 sen(x)]

**cos(2x)=cos²(x)-1

3 sen(x) + 4 cos(x) = (2cos²(x)-1) * [4 cos(x) − 3 sen(x)]

3 sen(x) + 4 cos(x) = 2cos²(x) * [4 cos(x) − 3 sen(x)] -[4 cos(x) − 3 sen(x)]

8 cos(x) = 2cos²(x) * [4 cos(x) − 3 sen(x)]

8 cos(x) = 8cos³(x) − 6cos²(x) * sen(x)

8cos³(x) − 6cos²(x) * sen(x) -8cos(x)=0

8cos³(x)-8cos(x) − 6cos²(x) * sen(x)=0

8cos(x)*[cos²(x)-1] − 6cos²(x) * sen(x)=0

8cos(x)*(-sen²(x)) − 6(1-sen²(x)) * sen(x)=0

-8cos(x)*sen²(x) − 6sen(x)+6sen³(x)=0

6sen(x)-6sen³(x)+8cos(x)*sen²(x)=0

2sen(x)*(3-3sen²(x)+4cos(x)*sen(x))=0

2sen(x)*(3(1-sen²(x))+4cos(x)*sen(x))=0

2sen(x)*(3cos²(x)+4cos(x)*sen(x))=0

2sen(x)*cos(x)*(3cos(x)+4sen(x))=0

sen(x)=0 ==>x=2pi*n  , x=pi+2pi*n

cos(x)=0 ==> x=pi/2 + 2pi*n , x=3pi/2+2pi*n

3cos(x)+4sen(x) =0 ==>tan(x)=-3/4 ==>x=arctan(-3/4)+pi*n

[tex]\begin{gathered}\Large\begin{array}{c} { 3 \sin(x) + 4 \cos(x) = \cos(2x) \cdot[4 \cos(x) - 3 \sin(x)] } \end{array}\end{gathered} \\ [/tex]

[tex]3 \sin(x) + 4 \cos(x) - \cos(2x) \left(4 \cos(x) - 3 \sin(x) \right) = 0 \\ \\ - \left( - 3 \sin( x ) + 4 \cos(x) \right)\cos(2x) + 3 \sin(x) + 4 \cos(x) [/tex]

Usando a identidade trigonométrica do arco duplo

Lembrando: [tex] \cos(2x) = 1 - 2 \sin^{2} (x) [/tex]

[tex] = 3 \sin(x) + 4 \cos(x) - (1 - 2 \sin^{2} (x) ) \\ ( - 3 \sin(x) + 4 \cos(x) ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex] = 3 \sin(x) + 4 \cos(x) + 3 \sin(x) - 4 \cos(x) - 6 \sin^{3} (x) + 8 \sin ^{2} (x) \cos(x) \\ \\ = 3 \sin(x) + 3 \sin(x) - 6 \sin^{3}(x) + 8 \sin^{2} (x) \cos(x) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex] = 6 \sin(x) - 6 \sin^{3} (x) + 8 \sin {}^{2} (x) \cos(x) \: \: \: \: \: \: \: \: \\ \\ = 6 \sin(x) - 6 \sin {}^{3} (x) + 8 \cos(x) \sin {}^{2} (x) \: \: \: \: \: \: \: \\ \\ = 6 \sin(x) - 6 \sin(x) \sin {}^{2} (x) + 8 \sin(x) \sin(x) [/tex]

[tex] = 3 \times 2 \sin(x) + 3 \times 2 \sin(x) \sin {}^{2} (x) + 4 \times 2 \sin(x) \sin(x) \\ \\ = 2 \sin(x) (3 - 3 \sin {}^{2} (x) + 4 \sin(x) \cos(x) ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex] = 2 \sin(x) (3 - 3 \sin {}^{2} (x) + 4 \sin(x) \cos(x) ) [/tex]

[tex] \sin(x) = 0 \\ ou \\ 3 - 3 \sin {}^{2} (x) + 4 \sin(x) \cos(x) = 0[/tex]

Vendo a tabela de periodicidade conclui-se que:

[tex]x = 0 + 2\pi n \\ x = \pi + 2\pi n[/tex]

[tex]x = 0 + 2\pi n \: \: ou \: \: 2\pi n \\ x = \pi + 2\pi n \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex]3 - 3 \sin {}^{2} (x) + 4 \sin(x) \cos(x) = 0 \\ \\ 3 - 3 \sin {}^{2} (x) + 4 \cos(x) \sin(x) \\ \\ = 3 - 3(1 - \cos {}^{2} (x) ) + 4 \cos(x) \sin(x) \\ \\ = 3 - 3 + 3 \cos {}^{2} (x) + 4 \cos(x) \sin(x) \\ \\ = 3 \cos {}^{2} (x) + 4 \cos(x) \sin(x) \\ \\ = 3 \cos(x) \cos(x) + 4 \sin(x) \cos(x) \\ \\ = \cos(x) (3 \cos(x) + 4 \sin(x)) = 0[/tex]

[tex] \cos(x) = 0 \\ ou \\ x = \frac{\pi}{2} + 2\pi n \: ,x = \frac{3\pi}{2} + 2\pi n[/tex]

[tex]3 \cos(x) + 4 \sin(x) = 0 \\ \\ 3 + 4 \tan(x) = 0 \\ \\ 4 \tan(x) = - 3 \\ \\ \frac{4 \tan(x ) }{4} = \frac{ - 3}{4} \\ \\ \tan(x) = - \frac{ 3}{4} \\ \\ x = arctan \left( - \frac{3}{4} \right) + \pi n[/tex]

Todas as soluções

[tex]x = 2\pi n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ x = \pi + 2\pi n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ x = \frac{\pi}{2} + 2\pi n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ x = \frac{3\pi}{2} + 2\pi n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ x = arctan \left( - \frac{3}{4} \right) + \pi n[/tex]

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