<s>[INST] Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ  = 5$$
Find 
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
 [/INST] 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,
$$
4\cos\theta-3\sin\theta=4\left(\frac35\right)-3\left(\frac45\right)=\Large\color{blue}0.
$$
</s>