11.16. Solve the equation

$$
3 \sin \varphi \cos \varphi + 4 \sin \varphi + 3 \cos^2 \varphi = 4 + \cos \varphi
$$
11.16. Let $x=\sin \varphi$ and $y=\cos \varphi$. The considered equation is equivalent to the system of equations

$$
3 x y+4 x+3 y^{2}-y-4=0, \quad x^{2}+y^{2}=1
$$

Given the condition $x^{2}+y^{2}=1$, the first equation is equivalent to the equation

$$
3 x y+4 x+3 y^{2}-y-4+\lambda\left(x^{2}+y^{2}-1\right)=0
$$

We will choose $\lambda$ such that the left-hand side can be represented as a product of two linear factors. According to problem 5.27, for this to be possible, the following equality must hold:

$$
\lambda(\lambda+3)(-4-\lambda)+2 \cdot \frac{3}{2} \cdot \frac{4}{2} \cdot\left(-\frac{1}{2}\right)=4(\lambda+3)+\frac{1}{4} \lambda+\frac{9}{4}(-4-\lambda) .
$$

It is not difficult to verify that this equation has a root $\lambda=-3$.

To obtain the factorization of the expression $3 x y-3 x^{2}+4 x-y-1$, first note that $3 x y-3 x^{2}=3 x(y-x)$. Next,

$$
(3 x+a)(y-x+b)=3 x y-3 x^{2}+(3 b-a) x+a y+a b
$$

Therefore, we need to set $a=-1$ and $b=1$. In the end, we get that the original equation is equivalent to the equation

$$
(3 \sin \varphi-1)(\cos \varphi-\sin \varphi+1)=0
$$