Question

resolver a equação cos( x-pi/4)= raiz de 3/2

Answer 1

$\mathbf{Sabemos\ que:}\\\\ \mathrm{\boxed{\mathrm{\cos{(a\pm b)=\cos{a}\cos{b}\mp\sin{a}\sin{b}}}}}\\\\\\ \mathbf{Ent\~{a}o:}$

$\mathrm{\cos{\bigg(x-\dfrac{\pi}{4}\bigg)}=\dfrac{\sqrt{3}}{2}\ \to\ \cos{x}\cos{\dfrac{\pi}{4}}+\sin{x}\sin{\dfrac{\pi}{4}}=\dfrac{\sqrt{3}}{2}}\\\\\\ \mathrm{\cos{x}\dfrac{\sqrt{2}}{2}+\sin{x}\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{3}}{2}\ \to\ \dfrac{\sqrt{2}}{2}\bigg(\cos{x}+\sin{x}\bigg)=\dfrac{\sqrt{3}}{2}}\\\\\\ \mathrm{\bigg(\cos{x}+\sin{x}\bigg)=\dfrac{\sqrt{3}}{\sqrt{2}}\ \to\ \bigg(\cos{x}+\sin{x}\bigg)^2=\bigg(\dfrac{\sqrt{3}}{\sqrt{2}}\bigg)^2}$

$\mathrm{\cos^2{x}+2\cos{x}\sin{x}+\sin^2{x}=\dfrac{3}{2}}\\\\\\ \mathbf{Lembrando\ que:}\\\\ \mathrm{\boxed{\mathrm{\cos^2{x}+\sin^2{x}=1}}\ \ e\ \ \boxed{\mathrm{\sin{2x}=2\sin{x}\cos{x}}}}\\\\\\ \mathrm{2\sin{x}\cos{x}+\cos^2{x}+\sin^2{x}=\dfrac{3}{2}\ \to\ \sin{2x}+1=\dfrac{3}{2}}\\\\\\ \mathrm{\sin{2x}=\dfrac{3}{2}-1\ \to\ \sin{2x}=\dfrac{1}{2}\ \to\ 2x=\arcsin{\bigg(\dfrac{1}{2}\bigg)}}$

$\mathrm{\mathbf{I.}\ 2x=\dfrac{\pi}{6}\ \to\ x=\dfrac{\frac{\pi}{6}}{2}\ \to\ \boxed{\mathrm{x=\dfrac{\pi}{12}}}}\\\\\\ \mathrm{\mathbf{II.}\ 2x=\dfrac{5\pi}{6}\ \to\ x=\dfrac{\frac{5\pi}{6}}{2}\ \to\ \boxed{\mathrm{x=\dfrac{5\pi}{12}}}}\\\\\\ \boxed{\boxed{\mathbf{\mathbb{S}=\bigg\{x\in\mathbb{R}\ \bigg|\ x=\dfrac{\pi}{12}\ \ ou\ \ x=\dfrac{5\pi}{12}\bigg\}}}}$