Express 3 x - 4 x in the form R(x + lambda), where R>0 and 0 < lambda < 90^.

$$R\cos(x + \lambda) = R(\cos x \cos \lambda - \sin x \sin \lambda)$$ $$R\cos(x + \lambda) = (R\cos \lambda)\cos x - (R\sin \lambda)\sin x$$

By comparing the coefficients of $\cos x$ and $\sin x$:

$$R\cos \lambda = 3 \quad \text{(Equation 1)}$$ $$R\sin \lambda = 4 \quad \text{(Equation 2)}$$

Square both equations and add them:

$$(R\cos \lambda)^2 + (R\sin \lambda)^2 = 3^2 + 4^2$$ $$R^2\cos^2 \lambda + R^2\sin^2 \lambda = 9 + 16$$ $$R^2(\cos^2 \lambda + \sin^2 \lambda) = 25$$

Since $\cos^2 \lambda + \sin^2 \lambda = 1$:

$$R^2 = 25$$

Given $R > 0$:

$$R = \sqrt{25} = 5$$

Divide Equation 2 by Equation 1:

$$\frac{R\sin \lambda}{R\cos \lambda} = \frac{4}{3}$$ $$\tan \lambda = \frac{4}{3}$$

Since $0 < \lambda < 90^\circ$, $\lambda$ is in the first quadrant.

$$\lambda = \arctan\left(\frac{4}{3}\right)$$ $$\lambda \approx 53.13^\circ$$

Substitute the values of $R$ and $\lambda$ into $R\cos(x + \lambda)$:

$$3\cos x - 4\sin x = 5\cos(x + 53.13^\circ)$$

The final answer is $5\cos(x + 53.13^\circ)$.