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 single idea that ties the whole dot point together is this: a trig equation almost never has one answer. Because $\sin$, $\cos$ and $\tan$ repeat, the equation $\sin x = \frac{1}{2}$ has infinitely many solutions across the real line, and a finite number inside any given interval. The marks live in finding every one of those, in order, and stopping exactly at the edges of the interval. The calculator hands you a single "principal" value; your job is to use the unit circle and the period to generate the rest.

The answer

Principal value and all solutions

The principal value is the standard "calculator" inverse:

• $x = \arcsin k$ gives $x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, valid for $-1 \le k \le 1$.
• $x = \arccos k$ gives $x \in [0, \pi]$, valid for $-1 \le k \le 1$.
• $x = \arctan k$ 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:

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

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

The whole method hangs on the unit circle. A point on the unit circle at angle $x$ (measured anticlockwise from the positive $x$-axis) has coordinates $(\cos x, \sin x)$. So $\sin x$ is the height of the point and $\cos x$ is its horizontal position. Solving $\sin x = k$ means "find every angle whose point sits at height $k$", which is exactly where the horizontal line $y = k$ cuts the circle: two points, symmetric about the $y$-axis. Solving $\cos x = k$ is where the vertical line $x = k$ cuts the circle: two points symmetric about the $x$-axis. This picture is why the symmetry formulas above take the shape they do.

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.

Finding all solutions on the unit circle, stage by stage

The whole procedure for a basic equation is best seen as a build-up on the unit circle. Take $\sin x = \frac{1}{2}$ on $0 \le x \le 2\pi$. Work through it in four stages.

Stage 1, recall the ASTC sign rule. Before placing anything, fix in mind which functions are positive in each quadrant. Starting from Q1 and going anticlockwise: All, Sine, Tan, Cos. Since $\frac{1}{2}$ is positive, the solutions to $\sin x = \frac{1}{2}$ must lie in the two quadrants where sine is positive, namely Q1 (all positive) and Q2 (sine positive).

Stage 2, place the principal value in Q1. The reference angle is $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ (this is also the principal value, since it lies in $[0, \pi]$). Draw the radius into Q1 at $\frac{\pi}{6}$; its tip sits on the line $y = \frac{1}{2}$. That is the first solution.

Stage 3, reflect into Q2. Sine is positive in Q2 as well, so reflect the Q1 radius across the $y$-axis. The reflected angle is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$, and its tip also sits on $y = \frac{1}{2}$. That is the second solution. (This is the "$\pi - \alpha$" rule for sine made visible.)

Stage 4, list both solutions and check the interval. The two radii give $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$, both inside $[0, 2\pi]$. There are no more, because sine is negative in Q3 and Q4, where the line $y = \frac{1}{2}$ does not meet the circle. The finished answer is $x = \frac{\pi}{6}, \frac{5\pi}{6}$.

The same four-stage discipline (sign rule, reference angle, place in the correct quadrants, then trim to the interval) solves every basic trig equation. Only the "correct quadrants" step changes: for a positive cosine the two radii are symmetric about the $x$-axis (Q1 and Q4); for a negative sine they drop into Q3 and Q4; and so on.

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 $\cos$

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 $\cos$.

How exam questions ask about solving trig equations

The wording tells you which technique is wanted. Learn to translate it:

• "Solve ... for $0 \le x \le 2\pi$" (or another interval). The interval is the instruction to give all solutions inside it, not just the principal value. Sweep both relevant quadrants, and check the endpoints.
• "Find the exact solutions" or a value like $\frac{\sqrt{3}}{2}$, $\frac{1}{2}$, $\frac{1}{\sqrt 2}$ appears. Use the exact-value triangle for the reference angle and leave the answer in terms of $\pi$, never a decimal.
• "Solve $\sin 2x = \dots$ / $\cos 3x = \dots$ / $\tan\frac{x}{2} = \dots$" (a multiple or fraction of $x$). Substitute $u$ for the inside, expand the interval to match, solve for $u$, then convert back. This is where careless candidates lose half the solutions.
• "Solve $2\cos^2 x + \sin x - 1 = 0$" (two different trig functions, one of them squared). Use a Pythagorean identity to reduce to a quadratic in a single function.
• "... $= 0$" with a product structure such as $\sin 2x = \cos x$. Bring everything to one side and factor; never divide by a trig function, or you discard the solutions where it is zero.
• "How many solutions does ... have in the interval?" You still solve it, but the count is the marked answer. Multiple-angle equations are the usual trick: $\sin 2x = k$ has up to four solutions in $[0, 2\pi]$, not two.
• "Hence solve ..." after a "prove that" part. The identity you just proved is the substitution to make; the word "hence" means you must use the previous result.