Question

Solve cos x +sqr root of 2 = -cos x for x over the interval 0,2pi

Answer 1

`\bf cos(x)+√(2)=-cos(x)\implies 2cos(x)+√(2)=0\implies 2cos(x)=-√(2) \\\\\\ cos(x)=-\cfrac{√(2)}{2}\implies \measuredangle x= \begin{cases} (3\pi )/(4)\\ (5\pi )/(4) \end{cases}`