Respuesta :

Answer:

Option A: [tex]\{0,\frac{\pi}{2},\frac{3\pi}{2}\}[/tex]

Step-by-step explanation:

[tex]sin^2(\theta)=2sin^2(\frac{\theta}{2}), [0,2\pi)[/tex]

[tex]sin^2(\theta)=2sin^2(\frac{\theta}{2})[/tex]

[tex]sin^2(\theta)=2sin(\frac{\theta}{2})sin(\frac{\theta}{2})[/tex]

[tex]sin^2(\theta)=2(\sqrt{\frac{1-cos(\theta)}{2}})(\sqrt{\frac{1-cos(\theta)}{2}})[/tex]

[tex]sin^2(\theta)=2(\frac{1-cos(\theta)}{2})[/tex]

[tex]sin^2(\theta)=\frac{2-2cos(\theta)}{2}[/tex]

[tex]sin^2(\theta)=1-cos(\theta)[/tex]

[tex]1-cos^2(\theta)=1-cos(\theta)[/tex]

[tex]-cos^2(\theta)=-cos(\theta)[/tex]

[tex]cos^2(\theta)=cos(\theta)[/tex]

[tex]cos^2(\theta)-cos(\theta)=0[/tex]

[tex]cos(\theta)[cos(\theta)-1]=0[/tex]

[tex]cos(\theta)=0[/tex]

[tex]\theta=\frac{\pi}{2},\frac{3\pi}{2}[/tex]

[tex]cos(\theta)-1=0[/tex]

[tex]cos(\theta)=1[/tex]

[tex]\theta=0[/tex]

Therefore, the solutions contained within the interval are [tex]\{0,\frac{\pi}{2},\frac{3\pi}{2}\}[/tex]

Helpful Tips:

Half-Angle Formula: [tex]sin(\frac{\theta}{2})=\pm\sqrt{\frac{1-cos(\theta)}{2}}[/tex]

Pythagorean Identity: [tex]sin^2(\theta)+cos^2(\theta)=1,sin^2(\theta)=1-cos^2(\theta),cos^2(\theta)=1-sin^2(\theta)[/tex]

loneguy

[tex]\sin^2 \theta = 2 \sin^2 \left(\dfrac{\theta}2 \right)\\\\\implies \sin^2 \theta = 1- \cos 2 \cdot \dfrac{\theta}2\\\\\implies \sin^2 \theta = 1- \cos \theta \\\\\implies 1-\cos^2 \theta = 1 - \cos \theta \\\\\implies -\cos^2 \theta - \cos \theta = 0\\\\\implies \cos^2 \theta - \cos \theta = 0\\\\\implies \cos \theta( \cos \theta -1) = 0\\\\\\\text{Now,}\\\\\cos \theta = 0\\\\\implies \theta = n\pi + \dfrac{\pi}2\\\\\text{For n = 0,1 and}~ [0.2\pi)\\\\[/tex]

[tex]\theta = \dfrac{\pi}2, \dfrac{3\pi}2[/tex]

[tex]\text{Again,} \\\\\cos \theta -1= 0\\\\\implies \cos \theta = 1\\\\\implies \theta = 2n\pi\\\\\text{For n= 0 and}~ [0,2\pi)\\\\\theta = 0\\\\\text{Combine solutions,}\\\\\theta = 0, \dfrac{\pi}2, \dfrac{3\pi}2[/tex]

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