Question

para 0 <_ x < 2pi

1) (cos x)elevado a 2 = 1

2) sen x = - raiz de 3/2

3) 3.tg x = - raiz de 3

4) 2.sen x.cos x + se n x = 0

5) cos 2 x - 4.cos x + 3 = 0

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Answer 1

$\boxed{\begin{array}{l}\tt 1)~\sf cos^2(x)=1\\\sf cos(x)=\pm\sqrt{1}\\\sf cos(x)=\pm1\\\sf cos(x) =1\\\sf x=0~ou~x=2\pi\\\sf cos(x)=-1\\\sf x=\pi\\\sf S=\{0,\pi, 2\pi\}\end{array}}$

$\boxed{\begin{array}{l}\tt 2)~\sf sen(x)=-\dfrac{\sqrt{3}}{2}\\\sf x=\dfrac{4\pi} {3}\\\sf ou~ x=\dfrac{5\pi}{3}\\\sf S=\bigg\{\dfrac{4\pi}{3}, \dfrac{5\pi}{3}\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\tt 3)~\sf 3\cdot tg(x) =-\sqrt{3}\\\sf tg(x)=-\dfrac{\sqrt{3}}{3} \\\sf x=\dfrac{5\pi}{6}\\\sf x=\dfrac{11\pi}{6}\\\sf S=\bigg\{\dfrac{5\pi}{6},\dfrac{11\pi}{6}\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\tt 4)~\sf2sen(x) cos(x) +sen(x)=0\\\sf sen(x)[2cos(x)+1]=0\\\sf sen(x)=0\\\sf x=0~ou~x=\pi~ou~x=2\pi\\\sf2cos(x)+1=0\\\sf 2cos(x)=-1\\\sf x=-\dfrac{1}{2}\\\sf x=\dfrac{2\pi} {3}\\\sf x=\dfrac{4\pi}{3}\\\sf S=\bigg\{0,\pi,2\pi,\dfrac{2\pi}{3},\dfrac{4\pi}{3}\bigg\} \end{array}}$

$\boxed{\begin{array}{l}\tt 5)~\sf cos(2x)-4cos(x)+3=0\\\sf2cos^2(x)-1-4cos(x)+3=0\\\sf 2cos^2(x)-4cos(x)+2=0\div2\\\sf cos^2(x) -2cos(x)+1=0\\\sf (cos(x)-1)^2=0\\\sf cos(x)-1=0\\\sf cos(x)=1\\\sf x=0~~ou~~x=2\pi\\\sf S=\{0,2\pi\}\end{array}}$