How to solve these trigonometric equations, using angle addition equations?

Question

$$4\sin(x)+7\cos(x)=6$$

where $0 \le x \le 360^{\circ}$

I put the equation into the form $a\sin(x)+b\cos(x)=R\sin(x+a)$, but after determining that $R\cos(a)=4, R\sin(a)=7$ and $R\sin(x+a)=6$, I don't know how to proceed. Any help?

Answer 1

Starting from $R=\sqrt{65}$, $a=\arcsin 7/\sqrt{65}$, we have

$$\sqrt{65}\sin(x+a)=6$$ $$\Rightarrow x=\arcsin \frac{6}{\sqrt{65}}-a=\arcsin \frac{6}{\sqrt{65}}-\arcsin \frac{7}{\sqrt{65}}$$

Using $$\arcsin u - \arcsin v=\arcsin (u\sqrt{1-v^2}-v\sqrt{1-u^2})$$ $$x=\arcsin\left(\frac{6}{\sqrt{65}}\cdot \frac{4}{\sqrt{65}}-\frac{7}{\sqrt{65}}\cdot \frac{\sqrt{65-6^2}}{\sqrt{65}}\right)$$ $$x=\arcsin \left(\frac{24-7\sqrt{29}}{65}\right)$$

Answer 2

HINT:

The R is related to

$\sqrt{4^2+7^2} = \sqrt{65},\text {so that} \sin a= \dfrac{7}{\sqrt{65}} ,\cos a= \dfrac{4}{\sqrt{65}}$

which is more convenient. Divide both sides by $\sqrt{65}$

$$ \sin ( x+ a) =\dfrac{6}{\sqrt{65}}$$

where

$$ \tan \alpha=\dfrac{7}{4}$$

Answer 3

$R\sin a = 7\\ R\sin a = 4\\ \tan a = \frac {\sin a}{\cos a} = \frac {7}{4}\\ a = \arctan \frac{7}{4}$

Now that we know $a$ what is $R$?

$\sin a = \frac {7}{R}\\ \cos a = \frac {4}{R}\\ \sin^2 a + \cos^2 a = 1\\ \left(\frac {7}{R}\right)^2 + \left(\frac {4}{R}\right)^2 = 1\\ \frac {49 + 16}{R^2} = 1\\ R^2 = 65\\ R = \sqrt {65}$

$\sqrt {65} \sin (x + \arctan\frac{7}{4}) = 6\\ x = \arcsin{\frac {6}{\sqrt{65}}} - \arctan \frac {7}{4}$

or

$x = \arcsin{\frac {6}{\sqrt{65}}} - \arcsin \frac {7}{\sqrt {65}}$

And for a general solution we will include $x+2n\pi$ and $x+\frac {4n-1}{2}\pi$

Answer 4

The simplest way to solve this equation is to write the l.h.s. in the form $$R\cos (x-\varphi)=6$$ Now $R=\sqrt{4^2+7^2}=\sqrt{65}$, and we have to solve (in radians) $$\begin{cases} \sin\varphi=\frac 4{\sqrt{65}}\\ \cos\varphi=\frac 7{\sqrt{65}} \end{cases} \iff \varphi=\arctan \tfrac 47$$ (because $\sin\varphi,\,\cos\varphi >0$).

Once $R$ and $\varphi$ have been determined, the equation becomes $$\cos(x-\varphi)=\frac 6R\iff x\equiv\varphi\pm \arccos\frac 6R\pmod{2\pi}.$$ The remaind to find the solutions which live in the interval $[0,2\pi]$, and translate it in degrees if you prefer a solution with this unit.