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The first four terms form the cubic Taylor polynomial about $x_0$:With a continuous fourth derivative nearby, Taylor's theorem gives a remainder $O((x-x_0{)}^4)$. Continuous derivatives through third order suffice for the weaker remainder $o((x-x_0{)}^3)$. These are local approximations; equality with an infinite Taylor series requires additional convergence information.

Use $2{\cos }^{2}x=1+\cos 2x$ to reduce the equation to $\sin 2x=\cos 2x$. Equivalently, $\sqrt{2}\sin (2x-\pi /4)=0$, so $x=\pi /8+k\pi /2$ for an integer $k$. Restricting to $[0,\pi ]$ gives

The known root gives a factor $x-1$ by the factor theorem. Polynomial division leaves $2x^2-7x+6=(2x-3)(x-2)$, soConsequently all the roots areThe three linear factors account for the full degree of the polynomial, so there are no further complex roots.