Daily math problems for August 18, 2026 — Trigonometry, Calculus, Probability & more

$\sin x = \cos x \iff \tan x = 1 \iff x = \pi/4 + k\pi$.

The denominator has higher degree, so the ratio tends to $0$.

Even outcomes: $\{2, 4, 6\}$, three out of six. Probability $= \dfrac{3}{6} = \dfrac{1}{2}$.

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

Every element is its own inverse: $A \triangle A = \emptyset$.

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

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

$$\det\begin{pmatrix}1&1&1\\1&m&1\\1&1&m\end{pmatrix}=1(m^2-1)-1(m-1)+1(1-m)$$

$\bullet$ 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 $m=1$.

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

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

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

$\bullet$ Case 1 ($|x|\geq 2$, so $x^2-4\geq 0$):

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

$\bullet$ Factor and solve:

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

$\bullet$ Case 2 ($|x|<2$, 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 $|x|<2$; $x=-2$ already appears in Case 1).

$\bullet$ Distinct valid solutions:

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

$\bullet$ Three real solutions.