If \(3\cos\theta+4\sin\theta=4\), where \(0 \le \theta < \dfrac{\pi}{2}\), then what is \(4\cos\theta+3\sin\theta\) equal to?
\(\dfrac{117}{25}\)
Let \(c=\cos\theta,\ s=\sin\theta\). From \(3c+4s=4\), \(c=\dfrac{4-4s}{3}\). Substituting in \(c^2+s^2=1\) gives \(25s^2-32s+7=0\), so \(s=1\) or \(s=\dfrac{7}{25}\). Since \(\theta<\dfrac{\pi}{2}\), \(s=1\) is rejected, so \(\sin\theta=\dfrac{7}{25}\) and \(\cos\theta=\dfrac{24}{25}\). Then \(4\cos\theta+3\sin\theta=\dfrac{96}{25}+\dfrac{21}{25}=\dfrac{117}{25}\). The correct option is (a).
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