Multiple Angle Equations

For $\sin u = k$ ($|k| \leq 1$), let $\alpha = \sin^{-1}|k|$ (the reference angle, always acute):

For $\cos u = k$, let $\alpha = \cos^{-1}|k|$:

For $\tan u = k$, let $\alpha = \tan^{-1}|k|$:

Once you have the general $u$-solutions, restrict to the widened domain and divide by $n$ to get $x$.

(nx) = k: expand domain to [0°, 360°n], find all u solutions, divide by n; Two branches for : u = + 360°k and u = 180° - + 360°k

Pause — copy the multiple-angle method into your book: let $u = nx$, expand domain to $[0°, 360°n]$, find all $u$-solutions, then divide each by $n$ to get $x$-solutions.

We just saw that $\sin(nx) = k$ requires expanding the domain to $[0°, 360°n]$ to capture all $n$ copies of the period. That raises a question: for $\sin 3x = \frac{1}{2}$ on $[0°, 360°]$, exactly how many $u$-solutions are there and what are the resulting $x$-values? This card answers it → domain becomes $[0°, 1080°]$, reference angle $30°$, giving 6 $u$-values and thus 6 $x$-solutions.

The function $y = \sin 3x$ completes three full cycles as $x$ runs from $0°$ to $360°$. Each full cycle of a sine wave crosses a horizontal line $y = k$ (with $|k| < 1$) twice. So there are $3 \times 2 = 6$ solutions.

More rigorously: letting $u = 3x$, the range for $u$ is $[0°, 1080°]$ — three full periods of $360°$ each. The two base solutions $u = \alpha$ and $u = 180° - \alpha$ repeat in each period:

Six values of $u$, giving six values of $x = u/3$.

For $\sin 3x = \frac{1}{2}$: $\alpha = 30°$.

$u = 30°, 150°, 390°, 510°, 750°, 870°$, so $x = 10°, 50°, 130°, 170°, 250°, 290°$.

3x = 1{2} → u = 3x, u [0°, 1080°], = 30°; u = 30°, 150°, 390°, 510°, 750°, 870°

Pause — copy the six-solution example into your book: $\sin 3x = \frac{1}{2}$, reference angle $30°$, domain $[0°, 1080°]$, giving $u = 30°, 150°, 390°, 510°, 750°, 870°$, so $x = 10°, 50°, 130°, 170°, 250°, 290°$.