Start with the equation:

\(7 \cos x + 5 = 2 \sin^2 x\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

\(7 \cos x + 5 = 2(1 - \cos^2 x)\)

Expand and rearrange:

\(7 \cos x + 5 = 2 - 2 \cos^2 x\)

\(2 \cos^2 x + 7 \cos x + 3 = 0\)

Factor the quadratic equation:

\((2 \cos x + 1)(\cos x + 3) = 0\)

Set each factor to zero:

\(2 \cos x + 1 = 0\) or \(\cos x + 3 = 0\)

\(\cos x = -\frac{1}{2}\) or \(\cos x = -3\)

Since \(\cos x = -3\) is not possible, solve \(\cos x = -\frac{1}{2}\):

\(x = 120^\circ, 240^\circ\)