Question

Resolva a equação
cos(2x) = sen(Π/4 - x) em [-pi,pi]

Answer 1

Resolver a equação trigonométrica

    $\mathsf{cos(2x)=sen\!\left(\dfrac{\pi}{4}-x\right)}$

para $\mathsf{x\in[-\pi,\,\pi].}$

Vamos trabalhar com a mesma função trigonométrica em ambos os membros. Reescreva o lado direito em termos de cosseno, usando a relação

    $\mathsf{sen(\theta)=cos\!\left(\dfrac{\pi}{2}-\theta\right)}$

para $\mathsf{\theta=\dfrac{\pi}{4}-x.}$  Dessa forma, a equação fica

    $\mathsf{\Longleftrightarrow\quad cos(2x)=cos\!\left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}-x\right)\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad cos(2x)=cos\!\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}+x\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad cos(2x)=cos\!\left(\dfrac{2\pi}{4}-\dfrac{\pi}{4}+x\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad cos(2x)=cos\!\left(\dfrac{\pi}{4}+x\right)}$

Agora temos uma igualdade entre cossenos. Use o seguinte resultado:

    $\mathsf{cos(\alpha)=\cos(\beta)\quad \Longleftrightarrow\quad \alpha=\pm\,\beta+k\cdot 2\pi,\qquad com~k\in\mathbb{Z}}$

para $\mathsf{\alpha=2x}$  e $\mathsf{\beta=\dfrac{\pi}{4}+x},$  e obtemos

    $\mathsf{\Longleftrightarrow\quad 2x=\pm\left(\dfrac{\pi}{4}+x\right)+k\cdot 2\pi}\\\\\\ \mathsf{\Longleftrightarrow\quad 2x=-\left(\dfrac{\pi}{4}+x\right)+k\cdot 2\pi\quad ou\quad 2x=\left(\dfrac{\pi}{4}+x\right)+k\cdot 2\pi}\\\\\\ \mathsf{\Longleftrightarrow\quad 2x=-\,\dfrac{\pi}{4}-x+k\cdot 2\pi\quad ou\quad 2x=\dfrac{\pi}{4}+x+k\cdot 2\pi}\\\\\\ \mathsf{\Longleftrightarrow\quad 2x+x=-\,\dfrac{\pi}{4}+k\cdot 2\pi\quad ou\quad 2x-x=\dfrac{\pi}{4}+k\cdot 2\pi}\\\\\\ \mathsf{\Longleftrightarrow\quad 3x=-\,\dfrac{\pi}{4}+k\cdot 2\pi\quad ou\quad x=\dfrac{\pi}{4}+k\cdot 2\pi}$

    $\mathsf{\Longleftrightarrow\quad \dfrac{12x}{4}=-\,\dfrac{\pi}{4}+\dfrac{8k\pi}{4}\quad ou\quad \dfrac{4x}{4}=\dfrac{\pi}{4}+\dfrac{8k\pi}{4}}\\\\\\ \mathsf{\Longleftrightarrow\quad 12x=-\,\pi+8k\pi\quad ou\quad 4x=\pi+8k\pi}\\\\\\ \mathsf{\Longleftrightarrow\quad 12x=\pi\cdot (-1+8k)\quad ou\quad 4x=\pi\cdot (1+8k)}\\\\\\ \mathsf{\Longleftrightarrow\quad x=\dfrac{\pi}{12}\cdot (8k-1)\quad ou\quad x=\dfrac{\pi}{4}\cdot (8k+1)}$

sendo k um número inteiro.

Mas para essa tarefa, queremos apenas as soluções no intervalo $\mathsf{[-\pi,\,\pi].}$  Para isso, atribua alguns valores inteiros para k.

•  Para $\mathsf{k\le -2,}$

    $\mathsf{\Longleftrightarrow\quad 8k\le -16}\\\\ \mathsf{\Longleftrightarrow\quad 8k-1\le -17\qquad e\qquad 8k+1\le -15}\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{\pi}{12}\cdot (8k-1)\le \dfrac{-17\pi}{12}<-\pi\qquad e\qquad \dfrac{\pi}{4}\cdot (8k+1)\le \dfrac{-15\pi}{4}<-\pi}$

Nesse caso, ambas as soluções não pertencem ao intervalo $\mathsf{[-\pi,\,\pi].}$

•  Para $\mathsf{k=-1,}$  encontramos

    $\mathsf{\Longrightarrow\quad x=\dfrac{\pi}{12}\cdot (8\cdot (-1)-1)\quad ou\quad x=\dfrac{\pi}{4}\cdot (8\cdot (-1)+1)}\\\\\\ \mathsf{\Longleftrightarrow\quad x=\dfrac{-9\pi}{12}\quad ou\quad x=\dfrac{-7\pi}{4}}\\\\\\ \mathsf{\Longleftrightarrow\quad x=-\,\dfrac{3\pi}{4}\quad ou\quad x=-\,\dfrac{7\pi}{4}<-\pi}$

Apenas a solução $\mathsf{x=-\,\dfrac{3\pi}{4}}$  pertence ao intervalo $\mathsf{[-\pi,\,\pi].}$

•  Para $\mathsf{k=0,}$  encontramos

    $\mathsf{\Longrightarrow\quad x=\dfrac{\pi}{12}\cdot (8\cdot 0-1)\quad ou\quad x=\dfrac{\pi}{4}\cdot (8\cdot 0+1)}\\\\\\ \mathsf{\Longleftrightarrow\quad x=-\,\dfrac{\pi}{12}\quad ou\quad x=\dfrac{\pi}{4}}$

Nesse caso, ambas pertencem ao intervalo $\mathsf{[-\pi,\,\pi].}$

•  Para $\mathsf{k=1,}$ encontramos

    $\mathsf{\Longrightarrow\quad x=\dfrac{\pi}{12}\cdot (8\cdot 1-1)\quad ou\quad x=\dfrac{\pi}{4}\cdot (8\cdot 1+1)}\\\\\\ \mathsf{\Longleftrightarrow\quad x=\dfrac{7\pi}{12}\quad ou\quad x=\dfrac{9\pi}{4}>\pi}$

Apenas a solução $\mathsf{x=\dfrac{7\pi}{12}}$  pertence ao intervalo $\mathsf{[-\pi,\,\pi].}$

•  Para $\mathsf{k\ge 2,}$

    $\mathsf{\Longleftrightarrow\quad 8k\ge 16}\\\\ \mathsf{\Longleftrightarrow\quad 8k-1\ge 15\quad e\quad 8k+1\ge 17}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{\pi}{12}\cdot (8k-1)\ge \dfrac{15\pi}{12}=\dfrac{5\pi}{4}>\pi\quad e\quad \dfrac{\pi}{4}\cdot (8k+1)\ge \dfrac{17\pi}{4}>\pi}$

Nesse caso, ambas as soluções não pertencem ao intervalo $\mathsf{[-\pi,\,\pi].}$

Portanto, reunindo as soluções encontradas, obtemos o conjunto solução para a equação dada:

    $\mathsf{\Longrightarrow\quad S=\left\{-\,\dfrac{3\pi}{4},\,-\,\dfrac{\pi}{12},\,\dfrac{\pi}{4},\,\dfrac{7\pi}{12}\right\}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)