Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$

HINT:

using Brahmagupta–Fibonacci identity

$$(4\sin\theta+3\cos\theta)^2+(4\cos\theta-3\sin\theta)^2=(4^2+3^2)(\sin^2\theta+\cos^2\theta)$$

Another approach:

Rewrite $$ \begin{align} 4\sin\theta +3\cos\theta &= 5\\ 4\sin\theta&=5-3\cos\theta.\tag1 \end{align} $$ Square $(1)$, yields $$ 16\sin^2\theta=25-30\cos\theta+9\cos^2\theta.\tag2 $$ Use identity $\sin^2\theta=1-\cos^2\theta$ and substitute to $(2)$. $$ \begin{align} 16(1-\cos^2\theta)&=25-30\cos\theta+9\cos^2\theta\\ 25\cos^2\theta-30\cos\theta+9&=0\\ (5\cos\theta-3)^2&=0\\ \cos\theta&=\frac35. \end{align} $$ Consequently, $\sin\theta=\dfrac45$. Thus, $$ 4\cos\theta-3\sin\theta=4\left(\frac35\right)-3\left(\frac45\right)=\Large\color{blue}0. $$

I learnt yet another way in High School (in the Netherlands).

$$4 \sin\theta + 3 \cos \theta = 5$$

Can be thought of as a vector dot product:

$$\binom{\cos\theta}{\sin\theta}\cdot\binom{3}{4}=5$$

Another way to write a dot product is $$\sqrt{\sin^2\theta+\cos^2\theta}\cdot\sqrt{4^2+3^2}\cdot \cos\phi = 5$$

where $\phi$ is the angle between the two vectors that make up the dot product.

This simplifies to $$\cos\phi=1$$

So $\phi=0$ or $\phi=\pi$.

Thus we know that the vector $\binom{\cos\theta}{\sin\theta}$ lies parallel to, or in the opposite direction of, $\binom{3}{4}$. In other words, we can write$$\tan\theta=\frac43$$

from which it follows that$$\theta = \arctan\left(\frac43\right) \mod \pi$$

Obviously this works for cases where $\phi\ne0$ - you just include whatever $\phi$ is.