# Solve the equation $7\sin\theta+3\cos\theta=4$ for all solutions in the interval $0°\leq\theta\leq360°$, giving $\theta$ to the nearest $0.1°$

Using the Addition Formulae

\begin{align} 7\sin\theta+3\cos\theta & = 4 \\ \Rightarrow \sin(\theta+23.2°) & = \frac{4}{\sqrt{58}} \\ \end{align}

\begin{align} \sin(\theta+23.2°) & = \frac{4}{\sqrt{58}} \\ \Rightarrow \theta+23.2° & = 31.7° \\ \Rightarrow \theta & = 31.7°-23.2° \\ \Rightarrow \theta & = 8.5° \\ \end{align} Therefore the solutions are, $$\theta=8.5°$$ \begin{align} \theta & = 180°-8.5° \\ \Rightarrow \theta & = 171.5° \\ \end{align}

The Mark Schemes solutions are, $$\theta=8.5°$$ $$\theta=125.1°$$ Where have I gone wrong? I've noticed that the second solution of $\theta$ is equal to, $$180-\left(\sin^{-1}\frac{4}{\sqrt{58}}\right)-23.2$$

but I cannot make any sense of this. Thanks

• You almost get the right answer. You mistake is: if $sin(θ+23.2°)=a$, then $θ+23.2°=arcsin(a) or θ+23.2°=180°-arcsin(a)$ – Huang May 27 '16 at 0:12