# 4\sin(x)+7\cos(x)=6 where 0 \leq x \leq 360^{\circ} I put the equation into

$$\displaystyle{4}{\sin{{\left({x}\right)}}}+{7}{\cos{{\left({x}\right)}}}={6}$$
where $$\displaystyle{0}\leq{x}\leq{360}^{{\circ}}$$
I put the equation into the form $$\displaystyle{a}{\sin{{\left({x}\right)}}}+{b}{\cos{{\left({x}\right)}}}={R}{\sin{{\left({x}+{a}\right)}}}$$, but after determining that $$\displaystyle{R}{\cos{{\left({a}\right)}}}={4},\ {R}{\sin{{\left({a}\right)}}}={7}$$ and $$\displaystyle{R}{\sin{{\left({x}+{a}\right)}}}={6}$$, I don't know how to proceed.

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ramirezhereva
Starting from $$\displaystyle{R}=\sqrt{{{66}}},{a}={\arcsin{{\frac{{{7}}}{{\sqrt{{{65}}}}}}}}$$ we have
$$\displaystyle\sqrt{{{65}}}{\sin{{\left({x}+{a}\right)}}}={6}$$
$$\displaystyle\Rightarrow{x}={\arcsin{{\frac{{{6}}}{{\sqrt{{{65}}}}}}}}-{a}={a}{r}{c}{i}{s}{n}{\frac{{{6}}}{{\sqrt{{{65}}}}}}-{\arcsin{{\frac{{{7}}}{{\sqrt{{{65}}}}}}}}$$
Using
$$\displaystyle{\arcsin{{u}}}-{\arcsin{{v}}}={\arcsin{{\left({u}\sqrt{{{1}-{v}^{{2}}}}-{v}\sqrt{{{1}-{u}^{{2}}}}\right)}}}$$
$$\displaystyle{x}={\arcsin{{\left({\frac{{{6}}}{{\sqrt{{{65}}}}}}\cdot{\frac{{{4}}}{{\sqrt{{{65}}}}}}-{\frac{{{7}}}{{\sqrt{{{65}}}}}}\cdot{\frac{{\sqrt{{{65}-{6}^{{2}}}}}}{{\sqrt{{{65}}}}}}\right)}}}$$
$$\displaystyle{x}={\arcsin{{\left({\frac{{{24}-{7}\sqrt{{{29}}}}}{{{65}}}}\right)}}}$$
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Vasquez

HINT:
NSK
The R is related to
so that $$\sin a=\frac{7}{\sqrt{65}}, \cos a=\frac{4}{\sqrt{65}}$$
which is more convenient. Divide both sides by $$\sqrt{65}$$
$$\sin(x+a)=\frac{6}{\sqrt{65}}$$
where
$$\tan \alpha=\frac74$$