Daily math problems for September 18, 2026 — Trigonometry, Logaritmi, Vectors & more

$\sin x(2\cos x-1)=0$. From $\sin x=0$: $x=0,\pi$. From $\cos x=\frac{1}{2}$: $x=\frac{\pi }{3},\frac{5\pi }{3}$. Total: $4$.

Combinăm: ${\log }_3(x(x-6))=3\Rightarrow x^2-6x=27\Rightarrow x^2-6x-27=0$. Rădăcinile sunt $x=9$ și $x=-3$. Domeniul cere $x>6$, deci $x=9$ (cealaltă este străină).

$\vec{u}\cdot \vec{v}=2\cdot 1+3\cdot (-4)=2-12=-10$.

Every element is its own inverse: $A\mathrm{△}A=\mathrm{\varnothing }$.

${\displaystyle \frac{\sin 3x}{\tan 5x}}={\displaystyle \frac{\sin 3x}{3x}}\cdot {\displaystyle \frac{5x}{\tan 5x}}\cdot {\displaystyle \frac{3}{5}}\to 1\cdot 1\cdot {\displaystyle \frac{3}{5}}={\displaystyle \frac{3}{5}}$.

$\bullet$ RHS must be non-negative, so $x\ge -2$.

$\bullet$ Case 1 ($\mathrm{\mid }x\mathrm{\mid }\ge 2$, so $x^2-4\ge 0$):

$$x^2-4=x+2\Rightarrow x^2-x-6=0$$

$\bullet$ Factor and solve:

$$(x-3)(x+2)=0\Rightarrow x=3,-2$$

$\bullet$ Case 2 ($\mathrm{\mid }x\mathrm{\mid }<2$, so $x^2-4<0$):

$$4-x^2=x+2\Rightarrow x^2+x-2=0$$

$\bullet$ Factor and solve:

$$(x+2)(x-1)=0\Rightarrow x=1$$

(only $x=1$ lies strictly inside $\mathrm{\mid }x\mathrm{\mid }<2$; $x=-2$ already appears in Case 1).

$\bullet$ Distinct valid solutions:

$$\{-2,1,3\}$$

$\bullet$ Three real solutions.

$b_4=10\sqrt{5}$. Since $\sqrt{5}\approx 2.236$, $b_4\approx 22.36$, so $\lfloor b_4\rfloor =22$.

$\bullet$ Compute the determinant of the coefficient matrix:

$$\det \left(\begin{array}{ccc}1& 1& 1\\ 1& m& 1\\ 1& 1& m\end{array}\right)=1(m^2-1)-1(m-1)+1(1-m)$$

$\bullet$ Simplify:

$$\det =m^2-2m+1=(m-1{)}^2$$

$\bullet$ Determinant vanishes only at $m=1$. At $m=1$ all three rows of the coefficient matrix become $(1,1,1)$ but the right-hand side is $(1,2,3{)}^T$:

$$x+y+z=1,2,3$$

$\bullet$ Three identical equations with different constants are inconsistent, so the system has no solution iff $m=1$.

$(2\sin x-1)(\sin x-1)=0$. From $\sin x=\frac{1}{2}$: $x=\frac{\pi }{6},\frac{5\pi }{6}$. From $\sin x=1$: $x=\frac{\pi }{2}$. Total: $3$.

$\widehat{2}\cdot \widehat{3}=\widehat{6}=\widehat{0}$ with $\widehat{2},\widehat{3}\mathrm{\ne }\widehat{0}$ — zero divisors. A field has no zero divisors, so ${\mathbb{Z}}_6$ is not a field. (Equivalently, $6$ is composite.)