Roots in trigonometry

By Vieta's we have that $\sin(x) + \cos(x) = -q$ and $\sin(x) \cos(x) = r$ , and thus

$\sin(x) + \cos(x) + \sin(x) \cos(x) = -q + r = -q + (q - 1) = -1 \Longrightarrow \sin(x) + \cos(x) + \sin(x) \cos(x) + 1 = 0 \Longrightarrow (\sin(x) + 1)(\cos(x) + 1) = 0$ ,

which is satisfied only if either $\sin(x) = -1 \Longrightarrow \cos(x) = 0$ or $\cos(x) = -1 \Longrightarrow \sin(x) = 0$ .

In either case this implies that $r = 0 \Longrightarrow q = 1$ , and thus $q + 3r = \boxed{1}$ .

If $r=q-1$ , then: $\begin{aligned}f(x)&=x^2+qx+r\\ &=x^2+qx+q-1\\ &=(x-1)(x+1)+q(x+1)\\ &=(x+1)(x-1+q)\end{aligned}$ Now, to find the roots we let $f(x)$ equal zero: $(x+1)(x-1+q)=0\\ x=-1,\qquad x=1-q$ But these two solutions must be in the form $\sin\theta,\cos\theta$ , and since $\sin^2\theta+\cos^2\theta=1$ : $(-1)^2+(1-q)^2=1\\ 1+1-2q+q^2=1\\ q^2-2q+1=0\\ (q-1)^2=0\\ q=1$ $\begin{aligned}\therefore q+3r&=q+3(q-1)\\ &=1+3(1-1)\\ &=\boxed{1}\end{aligned}$

We know that, for a general quadratic equation $\cdots$ $ax^{2} + bx + c$ , The sum of its roots is equal to $\frac{-b}{a}$ The product of the roots is equal to $\frac{c}{a}$ So, here we have $1x^{2} + qx + r$ , Therefore, the sum of roots i.e $\sin x + \cos x = \frac{-q}{1}\cdots$ (1) And, the product of roots i.e $\sin x . \cos x = \frac{r}{1}\cdots$ (2) Now, squaring on both sides of equation (1), we get $\sin^{2} x$ + $2\times {\sin x}\times {\cos x}$ + $\cos^{2} x$ $= q^{2}$ $\cdots$ (3), We know that, $\sin^{2} x + \cos^{2} x = 1$ , So, substituting equation (2) in equation (3), we get $1 + 2(q-1) = q^{2}$ $2(q - 1) = q^{2} - 1$ $2(q-1) = (q+1)(q-1)$ $q+1 = 2$ $\boxed{q=1}$ We need to find $q+3r$ It's given that $r=q-1$ So, $q+3r = 1 + 3(1-1)$ Hence, the answer is $\boxed{1}$

By Vieta's we know that $q = -(\sin x + \cos x)$ and $r = \sin x \ \cos x$ . Re-writing the given condition as $r + 1 = q$ we get

$\begin{array}{rcl} \sin x \ \cos x + 1 &=& -(\sin x + \cos x) \\ \\ \sin^2 x \ \cos^2 x + 2 \sin x \ \cos x + 1 &=& \sin^2 x + 2 \sin x \ \cos x\ + \cos^2 x \\ \\ \sin^2 x \ \cos^2 x &=& 0 \\ \\ \sin x = 0 &\text{OR}& \cos x = 0 \end{array}$

This yields $r = 0$ , which in turn yields $q = 1$ , so our answer is $q + 3r = \boxed{1}$