Solve trigonometric equations over a given interval using exact values, the unit circle, and identities to reduce to a single trig function

What this dot point is asking

NESA wants you to solve trigonometric equations on a specified interval, find all solutions (not just the principal value), handle multiple-angle equations with the right interval expansion, apply identities to reduce mixed equations to a single trig function, and recognise and solve quadratics in $\sin$ or $\cos$.

The answer

Principal value and all solutions

The principal value is the standard "calculator" inverse:

• IMATH_2 gives $x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, valid for $-1 \le k \le 1$.
• IMATH_5 gives $x \in [0, \pi]$, valid for $-1 \le k \le 1$.
• IMATH_8 gives $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, valid for all real $k$.

All other solutions come from symmetry plus periodicity. The full solution sets in $\mathbb{R}$ are:

• IMATH_12 : $x = \arcsin k + 2 k_1 \pi$ or $x = \pi - \arcsin k + 2 k_1 \pi$, $k_1 \in \mathbb{Z}$.
• IMATH_16 : $x = \pm \arccos k + 2 k_1 \pi$, $k_1 \in \mathbb{Z}$.
• IMATH_19 : $x = \arctan k + k_1 \pi$, $k_1 \in \mathbb{Z}$.

In an exam the interval is given (typically $[0, 2 \pi]$). Generate solutions from the formulas above and keep only those in the interval.

Quadrants and signs

The signs of the trig functions in each quadrant ("All Stations To Central" or ASTC):

For example, if $\cos x = -0.6$, $x$ is in Q2 or Q3.

Multiple-angle equations

For $\sin(b x) = k$ on $0 \le x \le L$, substitute $u = b x$. The interval for $u$ becomes $0 \le u \le b L$, which contains $b$ times as many solutions. Find all $u$ in that expanded interval, then divide each by $b$ to get the $x$ values.

This catches the easy-to-miss solutions that come from periodicity.

Reducing with identities

If an equation mixes trig functions, use identities to reduce to one. Common moves:

• Replace $\sin^2 x$ with $1 - \cos^2 x$ (or vice versa) using Pythagoras to get a polynomial in one function.
• Use $\sin 2x = 2 \sin x \cos x$ to expand a double-angle term, then factor.
• Use $\cos 2x = 1 - 2 \sin^2 x$ or $2 \cos^2 x - 1$ to convert between single and double angle forms.
• Divide both sides by $\cos x$ (provided $\cos x \neq 0$) to introduce $\tan x$.

Quadratics in $\sin$ or IMATH_54

An equation like $2 \sin^2 x - 3 \sin x + 1 = 0$ is a quadratic in $u = \sin x$. Factor or use the quadratic formula:

$(2 u - 1)(u - 1) = 0 \implies u = \frac{1}{2}$ or $u = 1$.

So $\sin x = \frac{1}{2}$ (solutions $x = \frac{\pi}{6}, \frac{5 \pi}{6}$) or $\sin x = 1$ (solution $x = \frac{\pi}{2}$), all in $[0, 2 \pi]$.

Reject any roots with $|u| > 1$ when $u$ stands for $\sin$ or $\cos$.

Worked examples

Basic sine equation

Solve $\sin x = -\frac{\sqrt{3}}{2}$ for $0 \le x \le 2 \pi$.

$\sin$ is negative in Q3 and Q4. Reference angle: $\arcsin\frac{\sqrt{3}}{2} = \frac{\pi}{3}$.

$x = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$ (Q3) or $x = 2\pi - \frac{\pi}{3} = \frac{5 \pi}{3}$ (Q4).

Cosine equation

Solve $\cos x = -\frac{1}{2}$ for $0 \le x \le 2 \pi$.

Principal value: $\arccos\left( -\frac{1}{2} \right) = \frac{2 \pi}{3}$.

Cosine has the same value at $x = -\frac{2 \pi}{3}$ which, shifted by $2 \pi$, gives $\frac{4 \pi}{3}$.

Solutions: $x = \frac{2 \pi}{3}, \frac{4 \pi}{3}$.

Multiple angle

Solve $\sin 3x = 0$ for $0 \le x \le \pi$.

Let $u = 3 x$, interval $0 \le u \le 3 \pi$.

$\sin u = 0$ at $u = 0, \pi, 2 \pi, 3 \pi$.

$x = 0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi$.

Reducing with Pythagoras

Solve $2 \cos^2 x + \sin x - 1 = 0$ for $0 \le x \le 2 \pi$.

Use $\cos^2 x = 1 - \sin^2 x$:

$2(1 - \sin^2 x) + \sin x - 1 = 0 \implies -2 \sin^2 x + \sin x + 1 = 0 \implies 2 \sin^2 x - \sin x - 1 = 0$.

Factor: $(2 \sin x + 1)(\sin x - 1) = 0$.

$\sin x = -\frac{1}{2}$: $x = \frac{7 \pi}{6}, \frac{11 \pi}{6}$.

$\sin x = 1$: $x = \frac{\pi}{2}$.

Three solutions in $[0, 2 \pi]$.

Using a double angle to factor

Solve $\sin 2 x = \cos x$ for $0 \le x \le 2 \pi$.

$2 \sin x \cos x - \cos x = 0 \implies \cos x (2 \sin x - 1) = 0$.

$\cos x = 0$: $x = \frac{\pi}{2}, \frac{3 \pi}{2}$.

$\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}, \frac{5 \pi}{6}$.

Four solutions: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$.

Common traps

Reporting only the principal value. A question with a given interval expects all solutions in that interval, not just one. Always sweep through quadrants and through periodic shifts.

Forgetting to expand the interval for multiple angles. If $u = 2 x$ and $x \in [0, 2\pi]$, then $u \in [0, 4\pi]$. Missing this halves the number of solutions found.

Dividing by $\cos x$ without checking. Dividing by $\cos x$ to introduce $\tan x$ loses the solutions where $\cos x = 0$. Either check those separately or factor instead of dividing.

Accepting $|\sin x| > 1$ or $|\cos x| > 1$. A quadratic in $\sin$ may produce a root outside $[-1, 1]$. Reject those, do not try to solve.

Wrong quadrants from the wrong identity. If $\sin x < 0$ and you "find" $x$ in Q1 or Q2, you have the wrong sign. Always use the quadrant rule (ASTC) to place the solution.

In one sentence

To solve a trig equation on an interval, isolate a single trig function (using identities if necessary), find the principal value, use the symmetry of the unit circle and the period to list every solution in the interval, and for multiple angles expand the interval first and then divide back at the end.