On sait que \(\displaystyle{\cos \left(x\right) = \dfrac{1+\sqrt 2 }{3}}\).
Donc l'équation devient :
\(\displaystyle{\left(\dfrac{1+\sqrt 2 }{3}\right)^2 + sin^2\left(x\right) = 1}\)
\(\displaystyle{\Leftrightarrow sin^2\left(x\right) = 1 - \left(\dfrac{1+\sqrt 2 }{3}\right)^2 }\)
\(\displaystyle{\Leftrightarrow sin^2\left(x\right) = 1 - \dfrac{1+2\sqrt 2 +2 }{9} }\)
\(\displaystyle{\Leftrightarrow sin^2\left(x\right) = \dfrac{9}{9} - \dfrac{3+2\sqrt 2 }{9} }\)
\(\displaystyle{\Leftrightarrow sin^2\left(x\right) = \dfrac{6-2\sqrt 2 }{9} }\)
On en déduit que :
\(\displaystyle{\sin\left(x\right) = \sqrt {\dfrac{6-2\sqrt 2 }{9} }}\) ou \(\displaystyle{\sin\left(x\right) = -\sqrt {\dfrac{6-2\sqrt 2 }{9} }}\)