aspifsGak5u

2022-01-03

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

ramirezhereva

Starting from $R=\sqrt{66},a=\mathrm{arcsin}\frac{7}{\sqrt{65}}$ we have
$\sqrt{65}\mathrm{sin}\left(x+a\right)=6$
$⇒x=\mathrm{arcsin}\frac{6}{\sqrt{65}}-a=arcisn\frac{6}{\sqrt{65}}-\mathrm{arcsin}\frac{7}{\sqrt{65}}$
Using
$\mathrm{arcsin}u-\mathrm{arcsin}v=\mathrm{arcsin}\left(u\sqrt{1-{v}^{2}}-v\sqrt{1-{u}^{2}}\right)$
$x=\mathrm{arcsin}\left(\frac{6}{\sqrt{65}}\cdot \frac{4}{\sqrt{65}}-\frac{7}{\sqrt{65}}\cdot \frac{\sqrt{65-{6}^{2}}}{\sqrt{65}}\right)$
$x=\mathrm{arcsin}\left(\frac{24-7\sqrt{29}}{65}\right)$

Vasquez

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

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