Over $0° \le x \le 360°$:

Because the graph rises and falls twice in that range, there are usually two solutions (occasionally one or none, depending on $k$) — stopping after finding just the calculator's answer is the single biggest cause of lost marks in this topic.

1. Find the principal (calculator) angle from the positive version of $k$: $\alpha = \sin^{-1}(|k|)$.
2. Use the sign of $k$ to pick the matching pair of quadrants:

For example, solving $5\sin x - 3 = 0$: $\sin x = 0.6$, so $\alpha = \sin^{-1}(0.6) = 36.9°$. Since $0.6>0$, the solutions are $x=36.9°$ and $x=180°-36.9°=143.1°$.

They come directly from the symmetry of the graphs, not from a rule to memorise blindly. The sine graph is a mirror image of itself about the vertical line $x=90°$, so any height reached at $x=\alpha$ is reached again at the mirrored position $x=180°-\alpha$. The cosine graph is symmetric about $x=180°$ instead, which is why its second solution pairs up as $360°-\alpha$ (a reflection about $x=180°$ measured the other way round the $360°$ range). Sketching the relevant graph and marking a horizontal line at height $k$ turns this from a memorised table into something read directly off a picture.

Solve $2\cos x + 1 = 0$: $\cos x = -0.5$. Find $\alpha$ from the positive value first, $\cos^{-1}(0.5)=60°$, then because $\cos x$ is negative, use the second/third-quadrant pair: $x = 180°-60°=120°$ and $x=180°+60°=240°$.

$\tan x$ repeats every $180°$, not $360°$, so its two solutions in range are always exactly $180°$ apart: $x=\alpha$ and $x=\alpha+180°$, regardless of the sign of $k$. This shorter repeat length is a direct consequence of the two branches of the tangent graph either side of each asymptote having identical shape.

Sketch $y=\cos x$ for $0°\le x\le 360°$ and use it to solve $\cos x = 0$.

Solution

Plot $(0,1),(90,0),(180,-1),(270,0),(360,1)$ and join with a smooth wave that peaks at the ends and troughs at $x=180°$. Reading where the curve crosses the $x$-axis gives $x=90°$ and $x=270°$ — matching the two points where $y=0$ in the table above.

Solve $4\sin x + 2 = 0$ for $0° \le x \le 360°$, giving all solutions to $1$ decimal place.

Solution

Since $\sin x$ is negative, the answer lies in the third and fourth quadrants. Find $\alpha$ from the positive value: $\alpha = \sin^{-1}(0.5) = 30°$.

Checking both against the range $0°$ to $360°$ confirms they're valid, and substituting $x=210°$ back into $\sin x$ gives $-0.5$ as expected — a quick way to catch an arithmetic slip before moving on.