Trigonometric equations and inequalities

Solving $\sin x = k$ starts with the inverse: $x_1 = \arcsin k$, a single angle in $[-\pi/2, \pi/2]$. But the sine wave hits the height k twice per cycle, once rising and once falling. The partner solution comes from symmetry: $x_2 = \pi - x_1$, the mirror across the vertical line at $\pi/2$.

Cosine's partner rule is different because its symmetry is different: from $x_1 = \arccos k$, the partner is $x_2 = 2\pi - x_1$ (equivalently $-x_1$), the mirror across the horizontal axis. Tangent needs no partner within its period at all: one solution per period, and the next lives a full $\pi$ away.

Worked on $[0, 2\pi)$: $\sin x = 1/2$ gives $x_1 = \pi/6$ and $x_2 = \pi - \pi/6 = 5\pi/6$. $\cos x = -1/2$ gives $x_1 = 2\pi/3$ and $x_2 = 2\pi - 2\pi/3 = 4\pi/3$. $\tan x = 1$ gives $\pi/4$, then $\pi/4 + \pi = 5\pi/4$. Two, two, and two.

Once the within-one-period solutions are in hand, periodicity mass-produces the rest: add $2\pi k$ (any integer k) to each sine or cosine solution, or $\pi k$ to each tangent solution. The general solution of $\sin x = 1/2$ is $x = \pi/6 + 2\pi k$ or $x = 5\pi/6 + 2\pi k$.

A stated domain then acts as a filter. Count carefully: a domain two periods wide holds twice as many solutions as one period. And special heights collapse the count: $\sin x = 1$ touches only once per cycle (at $\pi/2$), because the peak is where the two symmetric solutions merge. $\cos x = 1$ likewise gives only $x = 2\pi k$; writing $\pi k$ smuggles in $x = \pi$, where cosine is $-1$, not $1$.

Consider $\sin x \cos x = \sin x$ on $[0, 2\pi)$. Dividing both sides by $\sin x$ feels efficient and leaves $\cos x = 1$. But that division silently assumed $\sin x \neq 0$, and every input where $\sin x = 0$ was a legitimate candidate. The lawful move is to bring everything to one side and factor:

$$\sin x \cos x - \sin x = 0 \quad\Rightarrow\quad \sin x\,(\cos x - 1) = 0$$

Now each factor speaks: $\sin x = 0$ gives $x = 0, \pi$; $\cos x = 1$ gives $x = 0$. The complete set is $\{0, \pi\}$. Division would have delivered only $x = 0$ and quietly discarded $\pi$. The rule is absolute: never divide an equation by an expression that can be zero. Factoring costs one extra line and loses nothing.

Transformed arguments need a widened window instead. For $\sin 2x = 1/2$ on $[0, 2\pi)$, substitute $u = 2x$: as x runs through $[0, 2\pi)$, u runs through $[0, 4\pi)$, TWO full periods. Solve there ($u = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6$), then divide by 2: $x = \pi/12, 5\pi/12, 13\pi/12, 17\pi/12$. Four solutions, not two; solving in the original window and halving afterward drops half of them.

The AP exam works in radians, and a calculator left in degree mode poisons every inverse-trig step: asked for $\arcsin(0.4)$ it reports 23.6, a degree count masquerading as a radian answer (the radian value is about 0.412). The corrupted number then fails every subsequent computation. Check the mode before the first keystroke, and sanity-check outputs: an angle solution inside $[0, 2\pi)$ must be a number below about 6.28.

A trig inequality like $\sin x > 1/2$ is read off the graph or circle: find the boundary crossings ($\pi/6$ and $5\pi/6$), then ask which side of the wave sits above the line. The sine curve is above height 1/2 strictly BETWEEN the crossings: the solution on $[0, 2\pi)$ is the interval $(\pi/6, 5\pi/6)$, endpoints excluded because the inequality is strict. Solutions of trig inequalities are intervals, one per cycle, not isolated points.