Problem 1140

Solution. Method 1. Express $\cos x$ in terms of $\sin x$ (or vice versa), using the identity $\sin ^{2} x+\cos ^{2} x=1$, and square both sides of the equation (it is important to remember that squaring can introduce extraneous roots, so verification is necessary here).

Example. Solve the equation

$$3 \cos x+4 \sin x=5$$

We have:

$$\begin{gathered} 3 \cos x+4\left( \pm \sqrt{1-\cos ^{2} x}\right)=5, \quad \pm \sqrt{1-\cos ^{2} x}=\frac{5}{4}-\frac{3}{4} \cos x \\ 1-\cos ^{2} x=\left(\frac{5}{4}-\frac{3}{4} \cos x\right)^{2}, \quad(5 \cos x-3)^{2}=0, \quad \cos x=\frac{3}{5} \end{gathered}$$

We find $\sin x= \pm \frac{4}{5}$.

By verification, we ensure that the equation (2) is satisfied by such values of $x$ for which $\cos x=\frac{3}{5}$ and $\sin x=\frac{4}{5}$, i.e., $\operatorname{tg} x=\frac{4}{3}$.

Therefore, $x=\operatorname{arctg} \frac{4}{3}+\pi n, n \in \boldsymbol{Z}$ and $n$ is even, since the smallest positive period of $\sin$ and $\cos$ is $2 \pi$.

Method 2. Square the given equation and multiply the right side by $\sin ^{2} x+\cos ^{2} x$:

$$a^{2} \sin ^{2} x+2 a b \sin x \cos x+b^{2} \cos ^{2} x=c^{2}\left(\sin ^{2} x+\cos ^{2} x\right)$$

Divide both sides of this equation by $\cos ^{2} x$ (or by $\sin ^{2} x$), we get an equation equivalent to equation (3) (explain why!). However, squaring can introduce extraneous roots, so verification is also necessary here.

For example (2) we have: $9 \cos ^{2} x+24 \cos x \cdot \sin x+16 \sin ^{2} x=$

$$=25\left(\sin ^{2} x+\cos ^{2} x\right)$$

$9+24 \operatorname{tg} x+16 \operatorname{tg}^{2} x=25 \operatorname{tg}^{2} x+25$,

$$\operatorname{tg} x=\frac{4}{3}$$

!

Fig. 24

$$x=\operatorname{arctg} \frac{4}{3}+\pi k, k \in \boldsymbol{Z} \text { and } k \text { is even (see method } 1 \text { ). }$$

Method 3. Formulate a ready formula for solving equation (1). We will assume in equation (1) that $a \geqslant 0$ (if $a<0$, then it is sufficient to multiply both sides of the equation by -1).

Take a point $A(a ; b)$ on the circle centered at the origin (Fig. 24). Let the radius $O A$ form an angle $\varphi$ with the positive direction of the $O x$ axis, i.e., $\angle A O B=\varphi$. Since $a \geqslant 0$, then

$$-\frac{\pi}{2} \leqslant \varphi \leqslant \frac{\pi}{2}$$

From the triangle $A O B$ we have:

$$O A=r=\sqrt{a^{2}+b^{2}}, \quad \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}} \quad \text { and } \cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}} .$$

Therefore,

$$\begin{aligned} & a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \cdot\left(\frac{a}{\sqrt{a^{2}+b^{2}}} \sin x+\frac{b}{\sqrt{a^{2}+b^{2}}} \cos x\right)= \\ & =\sqrt{a^{2}+b^{2}} \cdot(\cos \varphi \cdot \sin x+\sin \varphi \cdot \cos x)=\sqrt{a^{2}+b^{2}} \cdot \sin (x+\varphi) \end{aligned}$$

Thus, equation (1) is equivalent to the equation

$$\begin{gathered} \sqrt{a^{2}+b^{2}} \cdot \sin (x+\varphi)=c, \text { or } \sin (x+\varphi)=\frac{c}{\sqrt{a^{2}+b^{2}}}, \\ \text { where } \varphi=\arcsin \frac{b}{\sqrt{a^{2}+b^{2}}} . \end{gathered}$$

From this,

$$x=-\arcsin \frac{b}{\sqrt{a^{2}+b^{2}}}+(-1)^{n} \arcsin \frac{c}{\sqrt{a^{2}+b^{2}}}+\pi n, \quad n \in Z$$

if $\left|\frac{c}{\sqrt{a^{2}+b^{2}}}\right| \leqslant 1$, or $c^{2} \leqslant a^{2}+b^{2}$.

For example (2) we have $a=4, b=3, c=5,5^{2} \leqslant 4^{2}+3^{2}$. By formula $(5)^{1}$

$$x=-\arcsin 0.6+(-1)^{n} \cdot \frac{\pi}{2}+\pi n, \quad n \in \boldsymbol{Z}$$