Daily math problems for August 17, 2026 — Trigonometry, Algebra, Probability & more

$x = \pi/6$ and $x = \pi - \pi/6 = 5\pi/6$. Two solutions.

$3 \cdot 5 = 15 = 2 \cdot 7 + 1$, so $\hat{3} \cdot \hat{5} = \hat{1}$.

A fair coin has two equally likely outcomes; the probability of heads is $\dfrac{1}{2}$.

$(1+i)^2 = 2i$, so $(1+i)^{10} = (2i)^5 = 32 \cdot i^5 = 32i$.

Matrices form a ring (with $0$ matrix and identity matrix), but multiplication is non-commutative for $n \ge 2$. It's a non-commutative ring with unity.

$a = 3$, $b = 4$, so asymptotes: $y = \pm \dfrac{4}{3}x$.

The graph is a semicircle; rotating about the $x$-axis sweeps out a sphere of radius $R$. (Volume $= \dfrac{4}{3}\pi R^3$.)

Midpoint: $(2, 0)$. Segment $AB$ is horizontal, so the bisector is vertical through $(2, 0)$: $x = 2$.

$\bullet$ Square the first matrix using $i^2=-1$:

$$\begin{pmatrix}3&2\\0&-1\end{pmatrix}^2=\begin{pmatrix}9&4\\0&1\end{pmatrix}$$

$\bullet$ Evaluate the entries of the second matrix: $\lim x^2/x!=0$, $\log_3 81=4$, $i^{2022}=-1$, $\tan 225°=1$:

$$\begin{pmatrix}0&4\\-1&1\end{pmatrix}$$

$\bullet$ Add and take the determinant:

$$\begin{pmatrix}9&8\\-1&2\end{pmatrix},\ \det=18-(-8)=26$$

$2\sin x \cos x - \sqrt{3}\cos x = 0 \Rightarrow \cos x\,(2\sin x - \sqrt{3}) = 0$.

$\bullet$ $\cos x = 0 \Rightarrow x \in \!\left\{\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right\}$

$\bullet$ $\sin x = \dfrac{\sqrt{3}}{2} \Rightarrow x \in \!\left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right\}$

Total: $4$ solutions on $[0, 2\pi)$.