Question

qual é o conjunto solução da equação :
$\mathtt{ \cos^2( \dfrac{3\pi}{2} - x )~=~\cos^2( \pi + x ) }$ que satisfazem a condição $\mathtt{ 0 \leq x \leq \pi }$ ??

A) $\huge{ \{ \dfrac{\pi}{6} ; \dfrac{5\pi}{6} \} }$

B) $\huge{ \{ \dfrac{\pi}{6} ; \dfrac{\pi}{3} \} }$

C) $\huge{ \{ \dfrac{\pi}{3} ; \dfrac{2\pi}{3} \} }$

​

Answer 1

$\mathsf{cos^2(\dfrac{3\pi}{2}-x)=cos^2(\pi+x)}\\\mathsf{|cos(\dfrac{3\pi}{2}-x)|=|cos(\pi+x)|}\\\mathsf{cos(\dfrac{3\pi}{2}-x)=cos(\pi+x)}\\\mathsf{-sen(x)=-cos(x)}\\\mathsf{sen(x)-cos(x)=0}\\\mathsf{(sen(x)-cos(x))^2=0^2}\\\mathsf{sen^2(x)-sen(2x)+cos^2(x)=0}\\\mathsf{1-sen(2x)=0\implies~sen(2x)=1}\\\mathsf{2x=\dfrac{\pi}{2}}\\\mathsf{4x=\pi}\\\mathsf{x=\dfrac{\pi}{4}}$

$\mathsf{cos(\dfrac{3\pi}{2}-x)=-cos(\pi+x)}\\\mathsf{-sen(x)=-1\cdot -cos(x)}\\\mathsf{-sen(x)=cos(x)}\\\mathsf{sen(x)+cos(x)=0}\\\mathsf{(sen(x)+cos(x))^2=0^2}\\\mathsf{sen(2x)+1=0}\\\mathsf{sen(2x)=-1}\\\mathsf{2x=\dfrac{3\pi}{2}}\\\mathsf{4x=3\pi}\\\mathsf{x=\dfrac{3\pi}{4}}$

Verificação:

Para $\mathsf{x=\dfrac{\pi}{4}:}$

$\mathsf{cos^2(\dfrac{3\pi}{2}-x)=cos^2(\dfrac{3\pi}{2}-\dfrac{\pi}{4})}\\\mathsf{cos^2(\dfrac{5\pi}{4})=\left(-\dfrac{\sqrt{2}}{2}\right)^2=\dfrac{1}{2}}$

$\mathsf{cos^2(\pi+\dfrac{\pi}{4})=cos^2(\dfrac{5\pi}{4})=\left(-\dfrac{\sqrt{2}}{2}\right)^2=\dfrac{1}{2}}$

Portanto

$\mathsf{\dfrac{\pi}{4}}$ é solução ✅

Para $\mathsf{x=\dfrac{3\pi}{4}:}$

$\mathsf{cos^2(\dfrac{3\pi}{2}-\dfrac{3\pi}{4})}\\\mathsf{cos^2(\dfrac{3\pi}{4})=\left(\dfrac{-\sqrt{2}}{2}\right)^2=\dfrac{1}{2}}$

$\mathsf{cos^2(\pi+\dfrac{3\pi}{4})=cos^2(\dfrac{7\pi}{4})=\left(\dfrac{\sqrt{2}}{2}\right)^2=\dfrac{1}{2}}$

Portanto

$\mathsf{\dfrac{3\pi}{4}}$ é solução ✅

$\huge\boxed{\boxed{\boxed{\boxed{\mathsf{S=\{\dfrac{\pi}{4},\dfrac{3\pi}{4}\}}}}}}$

$\dotfill$

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