Respuesta :
a)
The given equation is
[tex]5cot\theta-2=3cot\theta-2[/tex]Subtract 3cot(theta) from each side
[tex]\begin{gathered} 5cot\theta-3cot\theta-2=3cot\theta-3cot\theta-2 \\ 2cot\theta-2=-2 \end{gathered}[/tex]Add 2 to both sides
[tex]\begin{gathered} 2cot\theta-2+2=-2+2 \\ 2cot\theta=0 \end{gathered}[/tex]Divide both sides by 2
[tex]\begin{gathered} \frac{2cot\theta}{2}=\frac{0}{2} \\ \\ cot\theta=0 \end{gathered}[/tex]cot(theta) = 0 at theta = pi and theta = 3/2pi
[tex]\theta=\pi,\frac{3}{2}\pi[/tex]b)
[tex]2sin\theta=tan\theta[/tex]Change tan(theta) to sin(theta)/cos(theta)
[tex]\begin{gathered} tan\theta=\frac{sin\theta}{cos\theta} \\ \\ 2sin\theta=\frac{sin\theta}{cos\theta} \end{gathered}[/tex]Multiply both sides by cos(theta)
[tex]2sin\theta cos\theta=sin\theta[/tex]Subtract sin(theta) from both sides
[tex]2sin\theta cos\theta-sin\theta=0[/tex]Take sin(theta) as a common factor on the left side
[tex]sin\theta(2cos\theta-1)=0[/tex]Equate each factor by 0
[tex]\begin{gathered} sin\theta=0 \\ \theta=0,\pi \end{gathered}[/tex][tex]\begin{gathered} 2cos\theta-1=0 \\ 2cos\theta=1 \\ cos\theta=\frac{1}{2} \\ \theta=\frac{\pi}{3},\frac{5\pi}{3} \end{gathered}[/tex][tex]\theta=0,\frac{\pi}{3},\pi,\frac{5\pi}{3}[/tex]c)
[tex]3cos^2(\theta)-sin^2(\theta)=2[/tex]Change sin^2(theta) to 1 - cos^2(theta)
[tex]\begin{gathered} 3cos^2\theta-(1-cos^2\theta)=2 \\ 3cos^2\theta-1+cos^2\theta=2 \\ 4cos^2\theta-1=2 \\ 4cos^2\theta=3 \\ cos^2\theta=\frac{3}{4} \end{gathered}[/tex]Take a square root for each side
[tex]cos\theta=-\frac{\sqrt{3}}{2},cos\theta=\frac{\sqrt{3}}{2}[/tex]The values of theta are
[tex]\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}[/tex]